Physics Maha Marathon PDF

PHYSICS MAHA MARATHON  PRASHANT KIRAD Chapter Topics Numerical Mirror Formula...

PHYSICS MAHA MARATHON  PRASHANT KIRAD Chapter Topics Numerical Mirror Formula Lens Formula Chapter 9: Light Power of Lens All Ray Diagrams Snell’s Law Human Eye (Diagram) Chapter 10: Human Eye Defects (Myopia and Hypermetropia) Prism (Diagram + Concept) Numerical Series and Parallel Resistance R =ρ (l/A) Chapter 11: Electricity Power/ Heating effect Ohm’s Law Graph Calculating cost of Electricity of Appliance Fleming’s left hand rule (Numerical) Solenoid (Diagram) Chapter 12: Magnetic Effects Properties of Magnetic field lines Live wire, Neutral and earth wire.  PRASHANT KIRAD ELECTRICITY  PRASHANT KIRAD ELECTRICITY Conductors Semiconductors Insulators Have electrical conductivity between conductors and Allow electric current to flow easily. Do not allow electric current to flow easily. insulators. Conductivity can be altered by adding impurities or Contain free electrons. Lack free electrons. changing temperature. Examples: Copper, aluminum. Examples: Silicon, germanium. Examples: Rubber, plastic. ELECTRIC It is the property of matter that causes it to experience a force in an electric field. SI unit of charge = Coulomb (C). CHARGE: Electrons have negative charge (-e), and protons have positive charge (+e) e = 1.6 × 10⁻¹⁹ C). Properties: 1. Additivity of Charge :Total charge = sum of all charges on the body. 2. Charge is Conserved : Charge cannot be created or destroyed. 3. Charge is Invariant : Charge value remains the same, regardless of speed. Q = net charge 4. Quantization of Charge : Charge is a multiple of electron charge: n = no of electrons e = charge on an electrons QUANTISATION OF CHARGE: Q = ne. e = 1.6 × 10⁻¹⁹ C. According to charge quantization, any charged particle can have a charge equal to some integral number of e, i.e., Q = n e , where n=1, 2, 3,…  PRASHANT KIRAD Q. How many electrons are there in 1 coulomb of charge? ANSWER. Charge of one electron = 1.6 × 10⁻¹⁹ C Number of electrons (n) in 1 coulomb of charge: n = 1 ÷ (1.6 × 10⁻¹⁹) n = 6.25 × 10¹⁸ electrons So, 1 coulomb of charge contains approximately 6.25 × 10¹⁸ electrons.  PRASHANT KIRAD ELECTRIC CURRENT (I) Flow of electric charge through a conductor. Unit: Ampere (A) → 1 A = 1 C/s. One Ampere: When 1C of charge flows in 1 second then current is said to be 1A. Q I = current, Electrons flow from the negative terminal to the positive terminal of a battery. I= Q = charge, T t = time. Conventional current flows in the opposite direction, from the positive terminal to the negative terminal. The work done to move a unit positive charge W POTENTIAL DIFFERENCE between two points. V= Q Unit: Volt (V) → 1 V = 1 J/C. One Volt: 1 Joule of work done to move 1 unit positive charge between two points.  PRASHANT KIRAD Q. How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?  PRASHANT KIRAD Q. A current of 10 A flows through a conductor for two minutes. (i) Calculate the amount of charge passed through any area of cross-section of the conductor. (ii) If the charge of an electron is 1.6 × 10⁻¹⁹ C, then calculate the total number of electrons flowing.  PRASHANT KIRAD Q. A current of 10 A flows through a conductor for two minutes. (i) Calculate the amount of charge passed through any area of cross-section of the conductor. (ii) If the charge of an electron is 1.6 × 10⁻¹⁹ C, then calculate the total number of electrons flowing. Answer. (i) Calculation of charge : 𝑄 = 𝐼 × 𝑡 Given: I = 10 A, t = 2 minutes = 120 seconds Q = 10×120 = 1200 C (ii) Calculation of number of electrons : Q=n×e Where: Q = 1200 C e = 1.6 × 10⁻¹⁹ C, n = Q/e = 7.5 × 10²¹  PRASHANT KIRAD ELECTRIC CIRCUIT A continuous path for current flow, consisting of a power source, conductor, and load. Components: Electric devices. Source of electricity. Connecting wires and a switch to control the flow of current. A schematic diagram of an electric circuit comprising – cell, electric bulb, ammeter and plug key  PRASHANT KIRAD AMMETER & VOLTMETER Feature Ammeter Voltmeter Measures potential difference (voltage) Function Measures electric current in a circuit. between two points. Symbol Represented by A in a circuit diagram. Represented by V in a circuit diagram. Connection Connected in series with the circuit. Connected in parallel across the component. Resistance Has low resistance to allow current flow. Has high resistance to prevent current flow. Reading Unit Measured in amperes (A). Measured in volts (V).  PRASHANT KIRAD OHM’S LAW Current through a conductor is directly proportional to the potential difference across its ends, at a constant temperature When potential difference is 1 V V∝I and current through the circuit is 1 A, then resistance is 1 ohm. V = IR (i) Write the relation between resistance R and electrical resistivity ρ of the material of a conductor in the shape of a cylinder of length l and cross-sectional area A. Hence, derive the SI unit of electrical resistivity. Resistance: Property of a conductor that resists the flow of charges. Unit: Ohm (Ω). Factors Affecting Resistance: (i) Directly proportional to the length of conductor: R ∝ L (ii) Inversely proportional to the area of cross-section: R ∝ 1/A (iii) Directly proportional to the temperature: R ∝ Temperature (iv) Depends on nature of material: (⍴)  PRASHANT KIRAD Q. When a 4 V battery is connected across an unknown resistor there is a current of 100 mA in the circuit. The value of the resistance of the resistor is: a. 4 Ω b. 40 Ω c. 400 Ω d. 0.4 Ω  PRASHANT KIRAD Q. When a 4 V battery is connected across an unknown resistor there is a current of 100 mA in the circuit. The value of the resistance of the resistor is: a. 4 Ω b. 40 Ω c. 400 Ω d. 0.4 Ω Answer. Voltage (V) = 4 V Current (I) = 100 mA = 0.1 A Using Ohm’s Law: R=V/I R = 4 V / 0.1 A R = 40 Ω Correct answer: (b) 40 Ω  PRASHANT KIRAD Q. State and explain Ohm’s law. Define resistance and give its SI unit. What is meant by 1 ohm resistance? Draw V-I graph for an ohmic conductor and list its two important features Ans. It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically, V ∝ I V = IR ; where R is resistance of the conductor. Resistance : It is the property of a conductor to resist the How of charges through it. Its SI unit is ohm (Ω). If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 ohm (1 Ω). 1 ohm = 1volt/1ampere  PRASHANT KIRAD Q. A student plots V-I graphs for three samples of nichrome wire with resistances R₁, R₂, and R₃. Choose the correct statement from the following options: (a) R₁ = R₂ = R₃ (b) R₁ > R₂ > R₃ (c) R₃ > R₂ > R₁ (d) R₂ > R₁ > R₃  PRASHANT KIRAD Resistivity is a material's intrinsic property that measures its opposition to the flow of electric current. It is denoted by ρ. Resistivity does not change with change in length or area of cross-section but it changes with change in temperature. ⍴ = (R)A/L Resistance Resistivity Opposition to the flow of electric current in a substance. Resistance of a material with unit length and unit cross-sectional area. Extrinsic property. Intrinsic property. Depends on length and size of the conductor. Independent of length or size of the conductor. Unit: ohm (Ω). Unit: ohm-meter (Ω·m).  PRASHANT KIRAD Q. A wire of resistance 10 Ω is drawn out so that its length is thrice its original length. Find the resistance of the new wire.  PRASHANT KIRAD Q. A wire of resistance 10 Ω is drawn out so that its length is thrice its original length. Find the resistance of the new wire. Answer. Initial resistance, R₁ = 10 Ω New length, L₂ = 3L₁ Since resistance R is proportional to the square of the length when volume remains constant: R₂ = (new length / original length)² × R₁ R₂ = (3L₁ / L₁)² × 10 Ω R₂ = 9 × 10 Ω R₂ = 90 Ω  PRASHANT KIRAD Q. A wire has a resistance R and resistivity ρ. Answer the following: (a) What happens to the resistance if the length of the wire is doubled, keeping the cross-sectional area the same? (b) What happens to the resistance if the cross-sectional area is halved, keeping the length the same? (c) What happens to the resistivity if the material of the wire is changed?  PRASHANT KIRAD Q. A wire has a resistance R and resistivity ρ. Answer the following: (a) What happens to the resistance if the length of the wire is doubled, keeping the cross-sectional area the same? (b) What happens to the resistance if the cross-sectional area is halved, keeping the length the same? (c) What happens to the resistivity if the material of the wire is changed? Answer. (a) If the length is doubled: Since resistance is given by R = ρ (L/A), if L → 2L, then New resistance, R' = 2R (Resistance doubles). (b) If the cross-sectional area is halved: Since R = ρ (L/A), if A → A/2, then New resistance, R' = 2R (Resistance doubles). (c) If the material is changed: Resistivity ρ depends only on the material of the wire, not its dimensions. So, if the material is changed, ρ will also change accordingly.  PRASHANT KIRAD Q. Two wires A and B of the same material, having the same lengths and diameters 0.2 mm and 0.3 mm respectively, are connected one by one in a circuit. Which one of these two wires will offer more resistance to the flow of current in the circuit? Justify your answer.  PRASHANT KIRAD Q. Two wires A and B of the same material, having the same lengths and diameters 0.2 mm and 0.3 mm respectively, are connected one by one in a circuit. Which one of these two wires will offer more resistance to the flow of current in the circuit? Justify your answer. Ans. Wire A will offer more resistance. Justification: Resistance R∝l/A, where l is the length and A is the cross-sectional area. A thinner wire has a smaller cross-sectional area, leading to higher resistance. Wire A (0.2 mm diameter) has a smaller cross-sectional area compared to Wire B (0.3 mm diameter), so it offers higher resistance to the flow of current.  PRASHANT KIRAD Q. In case of four wires of same material, the resistance will be minimum if the diameter and length of the wire respectively are (a) D/2 and L/4 (b) D/4 and 4L (c) 2D and L (d) 4D and 2L  PRASHANT KIRAD When two or more resistors are connected end to end, the arrangement is called series combination. When resistors are connected in series, the total potential difference across the combination is the sum of the potential differences across each resistor. From the circuit: V = V₁ + V₂ + V₃ ------------- (1) Using Ohm's law, the potential difference across each resistor is given as: V₁ = IR₁, V₂ = IR₂, V₃ = IR₃ -----------------------(2) Substitute these into Eq. (1): V = IR₁ + IR₂ + IR₃ Current remains the same through Factor out I: V = I(R₁ + R₂ + R₃) -------------------------------(3) all resistors. For the equivalent single resistor Rₛ, using Ohm’s law: Voltage divides across resistors V = IRₛ --------------------------------------------(4) Comparing Eq. (3) and Eq. (4): based on resistance. Rₛ = R₁ + R₂ + R₃ Rₛ = R₁ + R₂ + R₃  PRASHANT KIRAD When two or more resistors are connected across multiple branches. When resistors are connected in parallel: The total current (I) is the sum of currents through each resistor: I = I₁ + I₂ + I₃ Using Ohm's law for the parallel combination: I = V / Rₚ Current remains the same through For each resistor: I₁ = V / R₁, I₂ = V / R₂, I₃ = V / R₃ all resistors. Voltage divides across resistors Substituting, we get: based on resistance. 1 / Rₚ = 1 / R₁ + 1 / R₂ + 1 / R₃ 1 / Rₚ = 1 / R₁ + 1 / R₂ + 1 / R₃  PRASHANT KIRAD Q. The combinations having equivalent resistance 1 is/are: (a) I and IV (b) Only IV (c) I and II (d) I, II, and III  PRASHANT KIRAD Q. What is the maximum resistance which can be made using five resistors each of 1/5 Ω? (a) 1/5 Ω (b) 10 Ω (c) 5 Ω (d) 1 Ω  PRASHANT KIRAD Q. In the given circuit diagram calculate (i) the total effective resistance of the circuit. (ii) the current through each resistor.  PRASHANT KIRAD Q. In the given circuit diagram calculate (i) the total effective resistance of the circuit. (ii) the current through each resistor. (i) Total Effective Resistance (Rₑq) Since all resistances are connected in parallel, the total resistance is given by: 1/Rₑq = 1/R₁ + 1/R₂ + 1/R₃ = 1/3 + 1/4 + 1/6 = (4+3+2)/12 = 9/12 So, Rₑq = 12/9 Ω = 4/3 Ω = 1.33 Ω (ii) Current through each resistor using Ohm’s Law Since the potential difference across each resistor connected in parallel is the same, we have: V₁ = V₂ = V₃ = 6V Applying Ohm’s law: V = IR For R₁: I₁ = V₁/R₁ = 6/3 = 2A For R₂: I₂ = V₂/R₂ = 6/4 = 1.5A For R₃: I₃ = V₃/R₃ = 6/6 = 1A Thus, the individual currents are: I₁ = 2A, I₂ = 1.5A, I₃ = 1A.  PRASHANT KIRAD Q. Three 2 Ω resistors A, B and C are connected in such a way that the total resistance of the combination is 3 Ω. Show the arrangement of the three resistors and justify your answer.  PRASHANT KIRAD Q. Three 2 Ω resistors A, B and C are connected in such a way that the total resistance of the combination is 3 Ω. Show the arrangement of the three resistors and justify your answer.  PRASHANT KIRAD Q. On the basis of this circuit, answer the following questions: (a) Find the total resistance between points A and B. (b) Find the resistance between points B and C. (c) (i) Calculate the current drawn from the battery when the key is closed.  PRASHANT KIRAD Q. On the basis of this circuit, answer the following questions: (a) Find the total resistance between points A and B. (b) Find the resistance between points B and C. (c) (i) Calculate the current drawn from the battery when the key is closed.  PRASHANT KIRAD Independent Functioning: Each device works independently. If one fails, others continue to work. Same Voltage: All components receive the same voltage as the source. Reduced Resistance: Adding more devices decreases the total resistance, allowing efficient current flow. Easy to Add Devices: New devices can be added without affecting others. Q. State Joule's law of heating. Derive the expression for the heat produced in a resistor. Joules Law of Heating ; Heat is proportional to the square of the current, resistance, and time. It states that the heat produced in a resistor is : (i) directly proportional to square of the current (I) (ii) directly proportional to resistance (R) for given current (iii) directly proportional to time (t) for which current flow through resistor.  PRASHANT KIRAD Electric Fuse: In any electrical instrument, due to sudden rise of current, the instrument gets burnt down which sometimes results in fire. A conducting wire with low melting point is connected in series with the circuit to avoid this type of accident. When the current rises, the wire melts due to excessive heating, thus breaking the electrical circuit. Electric Bulb: Electric bulb contains a thick metallic wire made up of tungsten metal. The metal is kept in an inert environment with a neutral gas or vacuum. When current flows through the tungsten wire, it becomes heated and emits light. Most of the electric power drawn in the circuit from the electrical source is dissipated in the form of heat and the rest is emitted in the form of light energy. Electric Heater/ Heating element: In an electric heater, high resistance nichrome wire is used as a coil. The coil is wound on grooves made up in ceramic material or china clay. When the current flows in the coil, it becomes heated, which is then used to heat cooking vessels. Elements used : Filament of the bulb - Tungsten Connecting wires - Copper Heating Elements - Nichrome Fuse wire - Sn - Pb alloy  PRASHANT KIRAD Q.(a) Why is tungsten used for making bulb filaments of incandescent lamps? (b) Name any two electric devices based on heating effect of electric current. (c) Why is an electric fuse used in household circuits? What is its principle? Answer: (a) (i) Tungsten is a strong metal and has high melting point (3380°C). (ii) It emits light at high temperatures (about 2500°C). (b) Electric laundry iron and electric heater are based on heating effect of electric current. (c) An electric fuse is used in household circuits because it protects electrical appliances from damage due to excessive current flow. Principle: It works on the heating effect of electric current—when excessive current flows through the fuse, the thin wire inside melts, breaking the circuit and preventing damage.  PRASHANT KIRAD The rate at which electric energy is consumed or dissipated in an electric circuit. Its S.I. unit is Watt (W) or J/s. It is denoted by P. W ΔE P= = t t  PRASHANT KIRAD Energy used by a circuit to allow current flow. It is the product of power and time, measured in watt-hours (Wh). Commercial Unit of Energy: One watt-hour is the energy used when 1 watt of power is consumed for 1 hour. The commercial unit of electric energy is the kilowatt-hour (kWh), also called a "unit." E=Pxt Joule = Watt x Second J = Ws E = Kilowatt hour  PRASHANT KIRAD Q. A coil in the heater consumes power P on passing current. If it is cut into halves and joined in parallel, it will consume power (a) P (b) P/2 (c) 2 P (d) 4 P Answer. (d) 4P  PRASHANT KIRAD Q. A coil in the heater consumes power P on passing current. If it is cut into halves and joined in parallel, it will consume power (a) P (b) P/2 (c) 2 P (d) 4 P  PRASHANT KIRAD Q.Two bulbs of 100W and 40W are connected in series.The current through the 100W bulb is 1A.The current through the 40W bulb will be: (a) 0.4 A (b) 0.6 A (c) 0.8 A (d) 1 A  PRASHANT KIRAD Q. (a) An electric bulb is rated at 200 V, 100W. What is its resistance? (b) Calculate the energy consumed by 3 such bulbs if they glow continuously for 10 hours for complete month of november.  PRASHANT KIRAD Q. (a) An electric bulb is rated at 200 V, 100W. What is its resistance? (b) Calculate the energy consumed by 3 such bulbs if they glow continuously for 10 hours for complete month of november.  PRASHANT KIRAD Q. A household uses the following appliances: 1 fan of 100 W for 10 hours/day 2 tube lights of 40 W each for 5 hours/day 1 refrigerator of 150 W running 24 hours/day Find the total energy consumption in kWh per month (30 days) and the cost if the rate is ₹6 per unit.  PRASHANT KIRAD Q. A household uses the following appliances: 1 fan of 100 W for 10 hours/day 2 tube lights of 40 W each for 5 hours/day 1 refrigerator of 150 W running 24 hours/day Find the total energy consumption in kWh per month (30 days) and the cost if the rate is ₹6 per unit. Answer. Energy used per day: Fan: (100 W × 10 h) ÷ 1000 = 1 kWh Tube lights: (2 × 40 W × 5 h) ÷ 1000 = 0.4 kWh Refrigerator: (150 W × 24 h) ÷ 1000 = 3.6 kWh Total per day = 1 + 0.4 + 3.6 = 5 kWh Total for 30 days = 5 × 30 = 150 kWh (units) Total cost = 150 × ₹6 = ₹900  PRASHANT KIRAD Q.Two identical resistors, each of resistance 15 Ω, are connected in (i) series, and (ii) parallel, in turn to a battery of 6 V. Calculate the ratio of the power consumed in the combination of resistors in each case.  PRASHANT KIRAD Q.Two identical resistors, each of resistance 15 Ω, are connected in (i) series, and (ii) parallel, in turn to a battery of 6 V. Calculate the ratio of the power consumed in the combination of resistors in each case. Answer.  PRASHANT KIRAD Q. B₁, B₂, and B₃ are three identical bulbs connected in a circuit as shown in Figure 12.8. When all three bulbs are glowing, the ammeter A records a current of 3A. Answer the following: 1. What happens to the glow of the other two bulbs if bulb B₁ gets fused? 2. What happens to the readings of ammeters A₁, A₂, A₃, and A if bulb B₂ gets fused? 3. Calculate the total power dissipated in the circuit when all three bulbs glow together.  PRASHANT KIRAD Q. B₁, B₂, and B₃ are three identical bulbs connected in a circuit as shown. When all three bulbs are glowing, the ammeter A records a current of 3A. Answer the following: 1. What happens to the glow of the other two bulbs if bulb B₁ gets fused? 2. What happens to the readings of ammeters A₁, A₂, A₃, and A if bulb B₂ gets fused? 3. Calculate the total power dissipated in the circuit when all three bulbs glow together. Step 1 Since all three bulbs are in parallel: Since R₁ = R₂ = R₃ = R: Using Ohm’s Law: Since Rₙₑₜ₁ = R/3, R=4.5Ω Step 3: Finding Current in Individual Branches Step 2: Finding Net Resistance and Current when B₂ Fuses Since B₂ is fused, no current flows through it: Now only B₁ and B₃ remain in parallel: A2=0AA_2 = 0AA2​=0A Total current in circuit: Total 2A divides equally between B₁ and B₃, since both have I=V/Rnet2 = 4.5/2.25 = 2A the same resistance: Ammeter A = 2A. A1=A3=2/2=1A  PRASHANT KIRAD Q. In a domestic circuit five LED bulbs are arranged as shown. The source voltage is 220 V and the power rating of each bulb is marked in the circuit diagram. Based on the following circuit diagram, answer the following questions: 1. State what happens when (i) key K1 is closed. (ii) key K2 is closed. 2. Find the current drawn by the bulb B when it glows. 3. Calculate (i) the resistance of bulb B, and (ii) total resistance of the combination of four bulbs B, C, D, and E. OR (c) What would happen to the glow of all the bulbs in the circuit when keys K1 and K2 both are closed and the bulb C suddenly gets fused? Give a reason to justify your PRASHANT KIRAD  PRASHANT KIRAD MAGNETIC EFFECTS OF ELECTRIC CURRENT  PRASHANT KIRAD Hans Christian Oersted (1820): Discovered that electric current deflects a compass needle, proving the link between electricity and magnetism. Magnet: is any substance that attracts iron or iron like substances. Properties of Bar Magnet: A freely suspended bar magnet aligns in the Earth's north-south direction. Attractive and Repulsive Forces: Like poles repel, opposite poles attract. Dipole Nature: Always has two poles (north and south); cutting the magnet creates smaller magnets, each with two poles. Creates a magnetic field around it where its effect can be felt. It retains its magnetic properties over time. Magnetic Field: is the area around a magnet in which the effect of magnetism is felt. Magnetic field lines are imaginary lines that show the strength and direction of a magnetic field. Horseshoe Bar magnet shaped  PRASHANT KIRAD Properties of Magnetic Filed Lines: Q. List the properties of magnetic field line Magnetic field lines start at the north pole and end at the south pole. Closer lines mean a stronger magnetic field (near poles). Field lines never cross each other. They form closed continuous curves. They show the direction of magnetic force. Q. State the right-hand thumb rule to find the direction of the magnetic field around a current-carrying conductor. The rule states that if a straight conductor carrying current is held in the right hand such that the thumb is pointed in the direction of the current, then the direction in which your fingers encircle the wire gives the direction of the magnetic lines of force around the wire Thumb = upwards, curled fingers = magnetic field (clockwise), the field direction = anticlockwise. Thumb = downwards, curled fingers = magnetic field (anticlockwise), the field direction = clockwise. concentric circles with their centers on the conductor  PRASHANT KIRAD circular pattern around the arms straight at the center of the loop electromagnet. A coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder. Outside the solenoid: North to South Inside the solenoid: South to North Factors: number of turns in the coil, amount of current flowing through it, radius of coil, Material of core of the solenoid. PRASHANT KIRAD  PRASHANT KIRAD Q. State Fleming's left-hand rule. Explain its application in understanding the direction of force on a current-carrying conductor in a magnetic field. When a current-carrying conductor is placed in an external magnetic field, the conductor experiences a force which is mutually perpendicular to both the Magnetic field and to the direction of the current flow. Stretch the thumb, forefinger, and middle finger of your left hand perpendicular to each other. Forefinger = Magnetic field direction, Middle finger = Current direction, Thumb = Force/motion direction. 1/100 second in India, i.e. the frequency of A.C in India is 50 Hz. Potential Difference in India: 220V at 50Hz. Alternating Current (AC) Direct Current (DC) AC can travel safely over long distances, even between cities. DC cannot travel long distances; it loses power. Frequency is 50 or 60 Hz, depending on the country. DC has zero frequency. Current direction reverses periodically. Current flows steadily in one direction. Cheaper then DC generation Expensive then AC generation  PRASHANT KIRAD Q. Imagine that you are sitting inside a chamber with your back to one wall. An electron beam which moving horizontally from the back wall towards the front wall, seems to be deflected by a strong magnetic field to your right side. What will be the direction of the magnetic field?  PRASHANT KIRAD Q. Imagine that you are sitting inside a chamber with your back to one wall. An electron beam which moving horizontally from the back wall towards the front wall, seems to be deflected by a strong magnetic field to your right side. What will be the direction of the magnetic field? Answer:Fleming’s Left Hand Rule can be used to calculate the magnetic field’s direction. The magnetic field will have a direction that is either upward or downward and perpendicular to the current and deflection axes. Because negatively charged electrons go from the back wall to the front wall, thus the direction of the current is from the front to the back wall. Rightward is where the magnetic force is directed. Therefore, it can be deduced using Fleming’s left-hand rule that the magnetic field inside the chamber is pointing downward.  PRASHANT KIRAD Q. Draw the pattern of magnetic field lines of a bar magnet and a current carrying solenoid. List two distinguishing features between the two fields of a bar magnet and a solenoid.  PRASHANT KIRAD Q. Draw the pattern of magnetic field lines of a bar magnet and a current carrying solenoid. List two distinguishing features between the two fields of a bar magnet and a solenoid. Answer. Magnetic field lines around a bar magnet. Magnetic field lines around a current carrying solenoid. Following are the distinguishing features between the two fields. (a) A bar magnet is a permanent magnet whereas solenoid is an electromagnet, therefore field produced by solenoid is temporary and stay till current flows through it. (b) Magnetic field produced by solenoid is more stronger than magnetic field of a bar magnet.  PRASHANT KIRAD Q. (a) State the purpose of the soft iron core used in making an electromagnet (b) List two ways of increasing the strength of an electromagnet if the material of the electromagnet is fixed.  PRASHANT KIRAD Q. (a) State the purpose of the soft iron core used in making an electromagnet (b) List two ways of increasing the strength of an electromagnet if the material of the electromagnet is fixed. Answer. (a)Electric bells and buzzers, loudspeakers, headphones, and other electrical devices all use electromagnets. The soft iron core that is inserted into an electromagnet makes the magnetic field produced stronger. Thus increasing the electromagnet’s strength in the process. (b) The strength of the electromagnet in use can be increased by (i) Increasing the current flowing through the coil. (ii) the number of turns in the coil may also be increased.  PRASHANT KIRAD Q. Give reason for the following (i) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid. (ii) The current carrying solenoid when suspended freely rests along a particular direction  PRASHANT KIRAD Q. Give reason for the following (i) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid. (ii) The current carrying solenoid when suspended freely rests along a particular direction. Answer:(i) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid because it behaves similar to that of a bar magnet and has a magnetic field line pattern similar to that of a bar magnet. Thus the ends of the straight solenoid behaves like poles of the magnet, where the converging end is the south pole and the diverging end is the north pole. (ii) The current carrying solenoid behaves similar to that of a bar magnet and when freely suspended aligns itself in the north-south direction.  PRASHANT KIRAD Power sockets (15A): For high-power appliances (geyser, fridge, AC). Normal sockets (5A): For low-power appliances (TV, bulbs, fans).  PRASHANT KIRAD Short Circuit : occurs when a live wire and a neutral wire come into direct contact, causing a sudden and large amount of current to flow in the circuit. Reasons: damage of insulation in power lines, fault in an electrical appliance. Overloading: If the total current drawn through a wire by the appliances connected to it exceeds the safety limit for that wire, it gets overheated. Electrical fuse: is a low melting point copper or other metal wire that breaks due to heat caused by overvoltage or high load to avoid short circuit or failure to the device. Q. (a) Why is it dangerous to touch the live wire rather than the neutral wire? (b) Why is a fuse important in household circuits? Answer. (a) Live wire is at 220V and neutral wire is at zero volt since the electric current flows from higher potential to lower potential, we can get an electric shock by touching live wire but that is not the case with neutral wire. (b)Fuse is an important safety device. It is used in series with any electrical appliance and protects it from short-circuiting and overloading.  PRASHANT KIRAD LIGHT  PRASHANT KIRAD The speed of light in vacuum is 3 × 10⁸ m/s. LIGHT Rectilinear Propagation of Light – Light travels in a straight line. A ray of light is the straight line along which light travels, and a bundle of light rays is called a beam of light. Q. State the laws of reflection and draw a labeled diagram to illustrate these laws. REFLECTION OF LIGHT The process of sending back light rays which falls on the surface of an object Incident Ray: The incoming ray of light that strikes the surface is called the incident ray. Reflected Ray: The ray that bounces off the surface is called the reflected ray. Normal: The imaginary line perpendicular to the surface at the point of incidence is called the normal. The Angle of incidence (∠i) = The angle of reflection (∠r) LAWS OF REFLECTION The incident ray, the reflected ray and the normal to the mirror at the point of incidence all lie in the same plane. Lateral Inversion – A phenomenon where an image appears reversed from left to right. This effect is commonly seen in mirrors, where your right hand appears as the left hand in the mirror image. Plane mirror: A smooth and polished surface that reflects light uniformly. The image obtained is virtual. The image is laterally inverted. The image is erect. The size of the image is the same as the size of the object. The distance between the image obtained from the mirror is the same as the distance between the object from the mirror.  PRASHANT KIRAD Terms Definition Pole (P) The center point of the reflecting surface of a spherical mirror. The center of the sphere of which the mirror's reflecting surface Centre of Curvature (C) forms a part. The radius of the sphere of which the mirror's reflecting surface Radius of Curvature (R) forms a part. R = 2f CONCAVE MIRROR The straight line passing through the pole and the center of Principal Axis curvature of the mirror. The point where parallel rays of light either converge or appear Principal Focus (F) to diverge after reflecting from the mirror. Focal Length (f) The distance between the pole and the principal focus. Aperture The diameter of the reflecting surface of the spherical mirror. CONVEX MIRROR  PRASHANT KIRAD Ray Diagrams (i) A ray parallel to principal axis will pass (iii) A ray passing through center of curvature will through focus after reflection. follow the same path back after reflection. (ii) A ray passing through the principal focus will (iv) Ray incident at pole is reflected back making become parallel to principal axis after reflection same angle with principal axis.  PRASHANT KIRAD Concave Mirror Convex Mirror  PRASHANT KIRAD Q. The image shows the path of incident rays to a concave mirror. Where would the reflected rays meet for the image formation to take place? (a) Behind the mirror (b) Between F and O (c) Between C and F (d) Beyond C  PRASHANT KIRAD Q. Draw ray diagrams for the following cases : (i) passing through centre of curvature of a concave mirror is incident on it. (ii) an object is placed between infinity and the pole of a convex mirror. (iii) object is placed between focus and pole of concave mirror  PRASHANT KIRAD Concave Mirror Uses of Concave Mirrors: Torches, Search-lights, and Vehicle Headlights: Shaving Mirrors Dentist's Mirrors Solar Furnaces Uses of Convex Mirrors: Convex Mirror Rear-View Mirrors in Vehicles: Preferred in vehicles as they provide erect but diminished images. Have a wider field of view due to their outward curve. Allow drivers to see a larger area compared to plane mirrors.  PRASHANT KIRAD Q. (a) Name the type of mirrors used in the design of solar furnaces. Explain how high temperatures are achieved by this device. (b) State the types of mirrors used for: (i) Headlights of vehicles (ii) Rear-view mirrors of motorcycles. Justify your answer in each case.  PRASHANT KIRAD Q. (a) Define the following terms in the context of spherical mirror: (i) Pole (ii) Centre of curvature (iii) Principal axis (iv) Principal focus (b) Consider the following diagram in which M is a mirror and P is an object and Q is its magnified image formed by the mirror. State the type of the mirror M and one characteristic property of the image Q.  PRASHANT KIRAD Q. (a) Define the following terms in the context of spherical mirror: (i) Pole (ii) Centre of curvature (iii) Principal axis (iv) Principal focus (b) Consider the following diagram in which M is a mirror and P is an object and Q is its magnified image formed by the mirror. State the type of the mirror M and one characteristic property of the image Q. Answer. (i) Pole: The middle point of the reflecting surface of a spherical mirror is called pole. The letter P represents pole, (ii) Centre of curvature: It is the centre of the sphere of glass of which the mirror is a part. The letter C represents the centre of curvature. (ii) Principal axis of a spherical mirror is the straight line joining the centre of curvature and pole of the mirroг. (iv)Principal focus: The mid-point of CP is called focus (F). It is the point on the principal axis of spherical mirror where all incident rays parallel to the principal axis meet or appear to diverge after reflection. (b) The mirror used in the given diagram is a concave spherical mirror. Image formed (Q) is virtual and magnified,  PRASHANT KIRAD Sign convention (i) The object is placed to the left of the mirror. (ii) All distances parallel to the principal axis are measured from the pole of the mirror. (iii) All distances measured in the direction of incident ray (along + X-axis) are taken as positive and those measured against the direction of incident ray (along – X-axis) are taken as negative. (iv) Distance measured perpendicular to and above the principal axis are taken as positive. Object distance = always positive (v) Distances measured perpendicular to and below the Focal length of concave mirror = Negative principal axis are taken as negative. Focal length of convex mirror = Positive  PRASHANT KIRAD MIRROR FORMULA The mirror formula relates the object distance (u), image distance (v), and focal length (f) of a spherical mirror. MAGNIFICATION It gives us information about the image in terms of how large or small is the image formed. h’ = positive (virtual images) h’ = negative (real images) m = negative (real) m = positive (virtual)  PRASHANT KIRAD Q. A student wants to obtain an erect image of an object using a concave mirror of 10 cm focal length. What will be the distance of the object from the mirror? (a) Less than 10 cm (b) 10 cm (c) Between 10 cm and 20 cm (d) More than 20 cm  PRASHANT KIRAD Q. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position of the image formed by the mirror.  PRASHANT KIRAD Q. The linear magnification produced by a spherical mirror is +3. Analyse this value and state the (i) type of mirror and (ii) position of the object with respect to the pole of the mirror.  PRASHANT KIRAD Q. The image of a candle flame placed at a distance of 30 cm from a mirror is formed on a screen placed in front of the mirror at a distance of 60 cm from it pole. What is the nature of the mirror? Find its focal length. If the height of the flame is 2.4 cm, find the height of its image. State whether the image formed is erect or inverted.  PRASHANT KIRAD Q. The image of a candle flame placed at a distance of 30 cm from a mirror is formed on a screen placed in front of the mirror at a distance of 60 cm from it pole. What is the nature of the mirror? Find its focal length. If the height of the flame is 2.4 cm, find the height of its image. State whether the image formed is erect or inverted. Answer. u = -30 cm, v = -60 cm, h = 2.4 cm Mirror Type: Since the image is real & on screen, the mirror is concave. Focal Length Calculation: Using 1/f = 1/v - 1/u: 1/f = -1/60 + 1/30 = 1/60 f = 60 cm Image Height: Using m = -v/u: m = 60/30 = 2 h' = 2 × 2.4 = 4.8 cm  PRASHANT KIRAD Q. An object 4 cm in height, is placed at 15 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image.  PRASHANT KIRAD Q. An object 4 cm in height, is placed at 15 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image. Answer. Given : object distance, u = -15 cm, object height, h = 4 cm, focal length f = -10 cm; Image distance, v = ? Using the mirror formula: 1/v + 1/u = 1/f Substituting the given values: 1/v + 1/(-15) = 1/(-10) 1/v = 1/15 - 1/10 v = -30 In order to obtain a sharp image of the object on the screen, screen should be placed at a distance of 30 cm in front of the mirror. m=h/h′​=−v​/u Substitute values: m= −(−30)/−15=2 h′=m×h=2×4= +8cm  PRASHANT KIRAD Q. Study the data given below showing the focal length of three concave mirrors A, B, and C and the respective distances of objects placed in front of the mirrors: Case Mirror Focal Length (cm) Object Distance (cm) 1 A 20 45 2 B 15 30 3 C 30 20 (i) In which one of the above cases will the mirror form a diminished image of the object? Justify your answer. (ii) List two properties of the image formed in Case 2. (iii) (A) What is the nature and size of the image formed by mirror C? Draw a ray diagram to justify your answer.  PRASHANT KIRAD The phenomenon of bending of ray of light when it enters from one medium to another. REFRACTION OF LIGHT The bending of a light ray during refraction occurs because of a change in the speed of light as it passes from one medium to another with a different refractive index. Incident Ray: The incoming ray of light in the first medium is called the incident ray. Reflected Ray: The ray that bends as it enters the second medium is called the refracted ray. Normal: The imaginary line perpendicular to the surface at the point of incidence is called the normal. CAUSES OF REFRACTION: When the light goes from air into water, it bends towards normal because there is a reduction in its speed. When the light goes from water to air, it bends away from normal because there is an increase in the speed of light. Rarer to denser medium (bends towards normal) Denser to rarer medium (bends away from normal) LAWS OF REFRACTION: The incident ray, refracted ray, and the normal to the interface of two media at the point of incidence all lie on the same plane. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. This is also known as Snell’s law of refraction Snell’s law of refraction.  PRASHANT KIRAD REFRACTIVE INDEX Measure of how much light is bent or refracted when it enters a new medium. It is denoted by the symbol "n." velocity of light in medium 1 n= velocity of light in medium 2 Speed of light in medium 1 Refractive index of medium 2 n 21 = with respect to medium 1 Speed of light in medium 2 Speed of light in air =c If medium 1 is vacuum or air, then the n m= refractive index of medium m is considered Speed of light in medium v with respect to vacuum. This is called the absolute refractive index of the medium. REFRACTION THROUGH A RECTANGULAR GLASS SLAB When an incident ray enters a glass slab from air, it bends towards the normal as it moves from a rarer to a denser medium. After passing through the slab, the refracted ray bends away from the normal as it exits back into air, forming an angle of emergence (e). The emergent ray remains parallel to the incident ray, with the perpendicular distance between them called lateral displacement. Light undergoes two refractions in a glass slab, causing this displacement.  PRASHANT KIRAD Q. A light ray passes from air into a glass slab. The angle of incidence is 30°, and the refractive index of glass with respect to air is 1.5. What is the angle of refraction? (A) 19.47° (B) 30° (C) 45° (D) 60°  PRASHANT KIRAD Q. The refractive index of glass is 1.5. The speed of light in vacuum is 3 × 10⁸ m/s. What is the speed of light in glass? A) 1 × 10⁸ m/s B) 2 × 10⁸ m/s C) 3 × 10⁸ m/s D) 4.5 × 10⁸ m/s  PRASHANT KIRAD Q. (a) Water has refractive index 1.33 and alcohol has refractive index 1.36. Which of the two medium is optically denser? Give reason for your answer. (b) Draw a ray diagram to show the path of a ray of light passing obliquely from water to alcohol. (c) State the relationship between angle of incidence and angle of refraction in the above case (d) In which medium will light travel faster? Explain with reason.  PRASHANT KIRAD Q. (a) Water has refractive index 1.33 and alcohol has refractive index 1.36. Which of the two medium is optically denser? Give reason for your answer. (b) Draw a ray diagram to show the path of a ray of light passing obliquely from water to alcohol. (c) State the relationship between angle of incidence and angle of refraction in the above case (d) In which medium will light travel faster? Explain with reason. Answer:(a) Here, alcohol is optically denser medium as its refractive index is higher than that of water. When we compare the two media, the one with larger refractive index is called the optically denser medium than the other as the speed of light is lower in this medium. (b) Since light is travelling from water (rarer medium) to alcohol (denser medium), it slows down and bends towards the normal. (d) Light travels faster in water because it has a lower refractive index.  PRASHANT KIRAD Q. A ray of light is incident as shown. If A, B and C are three different transparent media, then which among the following options is true for the given diagram? (a) ∠1 > ∠4 (b) ∠1 < ∠2 (c) ∠3 = ∠2 (d) ∠3 > ∠4  PRASHANT KIRAD LENSES A transparent material bound by two surfaces, of which one or both surfaces are spherical. Types of lenses: Convex (thicker in the middle, converging light rays) Concave (thinner in the middle, diverging light rays). Term Meaning Convex Lens A lens with two spherical surfaces bulging outwards, thicker in the middle than at the edges. (Converging Lens) Concave Lens A lens with two spherical surfaces curved inwards, thicker at the edges than at the middle. (Diverging Lens) Centre of Curvature (C, C1, C2) The center of the sphere from which the lens surface is a part. Principal Axis An imaginary straight line passing through the two centers of curvature of a lens. Optical Centre (O) The central point of a lens where a ray of light passes without deviation. Aperture The effective diameter of the circular outline of a spherical lens. Principal Focus (F, F1, F2) The point where rays of light parallel to the principal axis converge (convex) or appear to diverge (concave). Focal Length (f) The distance between the principal focus and the optical centre of a lens.  PRASHANT KIRAD Rules to obtain image (i) A ray of light from the object, (iii) A ray of light passing through the optical centre parallel to the principal axis of a lens (ii) A ray of light passing through a principal focus  PRASHANT KIRAD Convex Lens Concave Lens  PRASHANT KIRAD Convex Lens Uses of Convex Lens: overhead projector camera focus sunlight simple telescope projector microscope magnifying glasses Uses of Concave Lens: Concave Lens spy holes in the doors glasses some telescopes  PRASHANT KIRAD Q. A student wants to obtain magnified image of an object AB as on a Screen. Which one of the following arrangements shows the correct position of AB for him/her to be successful?  PRASHANT KIRAD Important Formulas of Lens: All measurements are taken from the optical centre of the lens. focal length of a convex lens = positive, and that of a concave lens = negative. Lens formula: Magnification: Power of Lens: The ability of a lens to converge or diverge the ray of light after refraction through it is called the power of the lens. It is defined as the reciprocal of focal length. SI unit = Dioptre (D) -1 1 dioptre is the power of a lens whose focal length is 1 metre. 1D = 1m. power of a convex lens = positive power of concave lens = negative.  PRASHANT KIRAD Q. The power of a lens is –2D. It means the lens is: (a) Convex with a focal length of 50 cm (b) Concave with a focal length of 50 cm (c) Convex with a focal length of 2 m (d) Concave with a focal length of 2 m  PRASHANT KIRAD Q. The focal length of four convex lens P, Q, R and S are 20 cm, 15 cm, 5 cm and 10 cm, respectively. The lens having the greatest power is (a) P (b) Q (c) R (d) S  PRASHANT KIRAD Q. An object is placed 30 cm from a convex lens, and its real image is formed on a screen 60 cm from the lens. a) Determine the focal length of the lens. b) If the object's height is 2.4 cm, calculate the height of the image. c) State whether the image formed is erect or inverted.  PRASHANT KIRAD Q. An object is placed 30 cm from a convex lens, and its real image is formed on a screen 60 cm from the lens. a) Determine the focal length of the lens. b) If the object's height is 2.4 cm, calculate the height of the image. c) State whether the image formed is erect or inverted. a) Focal Length Calculation: Using the lens formula: 1/f = 1/v - 1/u Substituting the values: 1/f = 1/60 - 1/(-30) f = 60/3 = 20 cm b) Image Height Calculation: m = v/u m = 60/(-30) m = -2 The negative sign indicates that the image is inverted. Image height (h') is calculated as: h' = m × h h' = -2 × 2.4 cm h' = -4.8 cm The negative sign indicates the image is inverted. c) Nature of the Image: The image formed is real, inverted, and magnified.  PRASHANT KIRAD Q. Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose. (i) State the nature of the lens and reason for its use. (ii) Where should the palmist place/hold the lens so as to have a real and magnified image of an object? (li) If the focal length of this lens is 10 cm, the lens is held at a distance of 5 cm from the palm, use lens formula to find the position and size of the image.  PRASHANT KIRAD Q. Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose. (i) State the nature of the lens and reason for its use. (ii) Where should the palmist place/hold the lens so as to have a real and magnified image of an object? (li) If the focal length of this lens is 10 cm, the lens is held at a distance of 5 cm from the palm, use lens formula to find the position and size of the image. Ans: (i) The lens used here is a convex lens and it is used as a magnifying glass because at close range, i.e., when the object is placed between optical centre and principal focus it forms an enlarged, virtual and erect image of the object. (ii) When this lens is placed such that the object is between the centre of curvature and the principal focus, the palmist obtain a real and magnified image. (iii) Given, focal length, f = 10 cm and u = -5 cm According to lens formula, Thus, the image will be formed at 10 cm on the same side of the palm and the size of the image will be enlarged.  PRASHANT KIRAD Q. One student uses a lens of focal length +50 cm and another of -50 cm. State the nature and find the power of each lens. Which of the two lenses will always give a virtual and diminished image irrespective of the position of the object?  PRASHANT KIRAD Q. One student uses a lens of focal length +50 cm and another of -50 cm. State the nature and find the power of each lens. Which of the two lenses will always give a virtual and diminished image irrespective of the position of the object?  PRASHANT KIRAD HUMAN EYE AND THE COLORFUL WORLD  PRASHANT KIRAD Sense organ for vision, located in the eye sockets of the skull; it helps us Human eye: Types of Neuron: see by detecting light and colors. Q. Draw a neat labelled diagram of the structure of the human eye and mention the functions of any five parts. Lens: Fibrous, jelly-like, convex; Ciliary Muscles: Hold and adjust Aqueous Humor: Clear fluid focuses light on the retina, the lens curvature for focus. between cornea and lens; creating a real, inverted image. maintains eye pressure and nourishes cornea and lens. Retina: Delicate membrane with light-sensitive cells. Pupil: Small opening in the iris; Rods: Detect light intensity. controls light entry into the eye. Cones: Detect primary colors. Iris: Ring-like, muscular tissue Optic Nerve: Transmits visual behind the cornea; determines eye information from the retina to the color and adjusts pupil size. brain. Cornea: Outermost transparent Vitreous Humor: Provides nutrients part; provides most light refraction. and maintains eye shape. Sclera: Tough, white outer covering of the eye; provides protection.  PRASHANT KIRAD Power of Accommodation The ability of the human eye to focus on objects at different distances by changing the focal length of the eye lens, controlled by ciliary muscles. Near point: Far point: Minimum distance for clear vision without Maximum distance seen clearly, strain, typically 25 cm for a normal eye. normally at infinity. When looking at a nearby object, the When looking at a distant object, ciliary muscles contract, making the the ciliary muscles relax, making lens thicker and decreasing its focal the lens thin and increasing its length. focal length. Gives a wider field of view. Enhances the ability to detect faint objects. Provides three dimensional view.  PRASHANT KIRAD Q. Define the term 'power of accommodation' of the human eye. How does the eye adjust to focus on objects at varying distances? Answer. The power of accommodation of the human eye is its ability to adjust the focal length of the eye lens to focus on objects at different distances. This is done by changing the curvature of the lens using the ciliary muscles. Adjustment Mechanism: For nearby objects: The ciliary muscles contract, making the lens thicker and increasing its converging power. For distant objects: The ciliary muscles relax, making the lens thinner and decreasing its converging power. This ability allows the eye to focus on objects at varying distances and form clear images on the retina.  PRASHANT KIRAD Defects of Vision & their Corrections Defect Description Cause Correction Image Formation - Excessive curvature - Can see nearby objects clearly Myopia (Near- of the eye lens. Concave lens of Image forms in but not distant ones. sightedness) - Elongation of the suitable power. front of the retina. - Far point is closer than infinity. eyeball. - Can see distant objects clearly - Focal length of eye Hypermetropia Convex lens of Image forms but not nearby ones. lens is too long. (Far-sightedness) appropriate power. behind the retina. - Near point is farther than 25 cm. - Eyeball is too small. - Weakening of ciliary Bifocal lenses - Difficulty seeing nearby objects Image formation muscles. (concave for Presbyopia due to aging. issues corrected - Reduced flexibility of distance, convex - Near point recedes with age. with bifocal lenses. the eye lens. for near vision). - Cannot focus on both horizontal - Irregularly shaped Image clarity Astigmatism and vertical lines simultaneously. cornea or distorted Cylindrical lens. varies based on - Objects clear in one plane only. lens. plane.  PRASHANT KIRAD cataract cloudy or blurry vision Causes: age related condition, weakening of eye muscles treatment: surgery. Refraction through a glass prism Prism: Transparent refracting medium. Structure: Two triangular bases, three rectangular lateral surfaces. Angle of Prism: Angle between two lateral faces. Angle of Deviation: Angle between incident and emergent rays.  PRASHANT KIRAD Q. A person is suffering from hypermetropia. List its two possible causes. Explain with a ray diagram how this defect is corrected.  PRASHANT KIRAD Q. A person is suffering from hypermetropia. List its two possible causes. Explain with a ray diagram how this defect is corrected. Answer: Hypermetropia (Farsightedness): A defect where a person can see distant objects clearly but cannot see nearby objects clearly. Causes: 1. Shortening of the eyeball. 2. Decreased curvature of the eye lens. It is corrected using a convex lens which converges the light rays before they enter the eye so that the image is formed on the retina.  PRASHANT KIRAD Q. Akshay, sitting in the last row in his class, could not see clearly the words written on the blackboard. When the teacher noticed it, he announced if any student sitting in the front row could volunteer to exchange his seat with Akshay. Salman immediately agreed to exchange his seat with Akshay. He could now see the words written on the blackboard clearly. The teacher thought it fit to send the message to Akshay’s parents advising them to get his eyesight checked. In the context of the above event, answer the following questions: (a) Which defect of vision is Akshay suffering from? Which type of lens is used to correct this defect? (b) State the values displayed by the teacher and Salman. (c) In your opinion, in what way can Akshay express his gratitude towards the teacher and Salman?  PRASHANT KIRAD Q. Riya is unable to see clearly the words written on the blackboard placed at a distance of approximately 3 m from her. (a) Name the defect of vision Riya is suffering from. (b) State two possible causes for this defect. (c) Explain why Riya is unable to see distinctly the objects placed beyond her far point. (d) With the help of Digarm - Describe the type of corrective lens used to restore proper vision and explain how this defect is corrected using the lens. (e) If the focal length of the corrective lens required for Riya is -3 m, calculate the power of the lens using the new Cartesian sign convention.  PRASHANT KIRAD Q. Riya is unable to see clearly the words written on the blackboard placed at a distance of approximately 3 m from her. (a) Name the defect of vision Riya is suffering from. (b) State two possible causes for this defect. (c) Explain why Riya is unable to see distinctly the objects placed beyond her far point. (d) With the help of Digarm - Describe the type of corrective lens used to restore proper vision and explain how this defect is corrected using the lens. (e) If the focal length of the corrective lens required for Riya is -3 m, calculate the power of the lens using the new Cartesian sign convention. (a) Riya is suffering from myopia (short-sightedness). (b) Possible causes of myopia: 1. Elongation of the eyeball from front to back. 2. Excessive curvature of the eye’s cornea or lens. (c) In myopia, the image of a distant object is formed in front of the retina instead of directly on it. This happens because the light rays converge too early due to the elongated eyeball or excessively curved lens. (d) A concave lens (diverging lens) is used to correct myopia. The concave lens diverges the incoming light rays before they enter the eye, effectively extending the far point of the person to infinity. This ensures the image forms directly on the retina. (e) focal length: f=−3m Using the formula for power: P=1/f (in meters)orP=1/−3=−0.33 D Hence, the power is – 0.33 D.  PRASHANT KIRAD Dispersion of White Light: Splitting of white light into seven colors when passing through a prism. Spectrum: The band of seven colors formed. Color Sequence: VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red). Causes: Varying refraction indices of different colours. wavelength of light when passing through transparent medium like prism. ELECTROMAGNETIC SPECTRUM The visible light range is from 400 nm to 700 nm in wavelength. Violet: 400 nm (shortest wavelength), Red: 700 nm (longest wavelength) Wavelength ∝ Velocity ∝ 1/Deviation In vacuum and air, all colors of light travel at the same speed. In other media, they travel at different speeds, causing dispersion. Red light travels fastest and deviates least, while violet light travels slowest and deviates most. Seven coloured lights of the spectrum." can be recombined to give back Recombination of white light white light by passing two prism one by upside down. Newton’s Experiment: Used a second inverted prism to recombine the spectrum into white light, proving sunlight is made up of seven colors. White Light: Any light producing a similar spectrum to sunlight is called white light.  PRASHANT KIRAD Q. (a) With the help of labelled ray diagram show the path followed by a narrow beam of monochromatic light when it passes through a glass prism. (b) What would happen if this beam is replaced by a narrow beam of white light?  PRASHANT KIRAD Q. (a) With the help of labelled ray diagram show the path followed by a narrow beam of monochromatic light when it passes through a glass prism. (b) What would happen if this beam is replaced by a narrow beam of white light? Here, in the figure, ∠D is the angle of deviation of the given monochromatic light by the glass prism. (b) If AO were a ray of white light, then on screen BC, a spectrum will be observed, consisting of seven colours arranged from bottom to top as follows. Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR)  PRASHANT KIRAD Q. What is dispersion of light? Explain with a diagram how a prism disperses white light into its constituent colors.  PRASHANT KIRAD Q. What is dispersion of light? Explain with a diagram how a prism disperses white light into its constituent colors. Answer: Dispersion: The splitting of white light into its seven colors (VIBGYOR) when passing through a prism. Cause: Different colors of light have different wavelengths and bend by different angles when passing through a glass prism.  PRASHANT KIRAD Q. How will you use two identical glass prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw and label the ray diagram.  PRASHANT KIRAD Q. How will you use two identical glass prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw and label the ray diagram. Answer: Newton was the first to use a glass prism to obtain the spectrum of a white light. He then placed a second identical prism in an inverted position with respect to the first prism. This allowed all the colours of the white light to pass through the second prism combining to form a white light emerging from the other side of the second prism. This made him believe that white light was composed of different colours.  PRASHANT KIRAD A natural spectrum appearing in the sky after a rain shower, caused by Natural spectrum: Rainbow : the dispersion of sunlight by tiny water droplets in the atmosphere. Refraction of Sunlight Dispersion into Colors Internal Reflection Refraction Again Mechanism: Water droplets act like prisms, refracting and dispersing sunlight, reflecting it internally, and refracting it again. Color Sequence: Red at the top, violet at the bottom. Formation Direction: Always opposite to the sun. Atmospheric Refraction The refraction of light caused by the Earth‘s atmosphere (having air layers of varying optical densities) Stars Twinkle Advanced Sunrise & Delayed Sunset: Why Planets Don’t Twinkle Caused by Sun appears ~2 minutes before sunrise and Planets are closer and seen as atmospheric after sunset due to atmospheric refraction. The extended sources, averaging out the refraction; starlight Sun's disc also appears flattened at these times. light variations and reducing the bends as it enters twinkling effect. Earth's atmosphere, causing stars to appear to change position and flicker.  PRASHANT KIRAD Q. Describe how a rainbow is formed when sunlight passes through raindrops. Include the roles of refraction, dispersion, and reflection in your answer.  PRASHANT KIRAD Q. Describe how a rainbow is formed when sunlight passes through raindrops. Include the roles of refraction, dispersion, and reflection in your answer. Answer. A rainbow is formed when sunlight passes through raindrops in the atmosphere, involving three key processes: refraction, dispersion, and reflection. 1. Refraction: When sunlight enters a raindrop, it slows down and bends due to the change in medium from air to water. This bending of light is known as refraction. 2. Dispersion: As the light is refracted, it also disperses into its constituent colors (red, orange, yellow, green, blue, indigo, violet) because different colors of light have different wavelengths and refract at slightly different angles. This separation of colors is called dispersion. 3. Reflection: After dispersion, the light reflects off the inner surface of the raindrop. This internal reflection causes the light to bounce back toward the outside of the raindrop. 4. Emergence: Finally, as the light exits the raindrop, it undergoes a second refraction. This further bending of the light enhances the separation of colors, allowing us to see a spectrum of colors in the form of a rainbow.  PRASHANT KIRAD Scattering of Light Scattering of light occurs when light is absorbed by particles and then re-emitted in different directions. Red Sun at Sunrise/Sunset Blue Sky During sunrise and sunset, sunlight travels a longer due to the scattering of sunlight by small air distance through the atmosphere. Blue light is molecules and fine particles. Blue light, having a scattered away, while red light, with a longer shorter wavelength, scatters more than red light, wavelength, reaches the observer's eyes, making the making the sky appear blue. sun appear red. Tyndall Effect: Light scatters when it strikes particles in a colloid, making the light path visible. Seen in sunlight passing through mist or a dense forest canopy. Smaller particles scatter blue light, while larger particles scatter red light.  PRASHANT KIRAD Q. Explain the Tyndall effect. (a) Why does the sky appear blue during the day and (b) reddish at sunrise and sunset? Answer. The Tyndall effect is the scattering of light by tiny particles in a medium, making the path of light visible. It is more effective for shorter wavelengths (blue) than longer wavelengths (red). (a) During the day, shorter wavelengths (blue) scatter more in all directions due to atmospheric particles, making the sky appear blue. (b) At sunrise and sunset, sunlight travels a longer path through the atmosphere. Most of the blue light is scattered out, leaving red and orange hues visible. PRASHANT KIRAD