Physics Past Paper PDF

 =1𝑚 =4𝑚 = 0.25 𝑚/𝑠 = 1.75 𝑚/𝑠 = 0.5 𝑚/𝑠² = 0.5 𝑚/𝑠² =1𝑚 = 𝟐. 𝟓 𝑚 = 0.25 𝑚/𝑠 = 1.25 𝑚/𝑠 = 0.5 𝑚/𝑠² = 0.5 𝑚/𝑠² =1𝑚 =4𝑚 = 0.25 𝑚/𝑠 = 1.75 𝑚/𝑠 = 0.5 𝑚/𝑠² = 0.5 𝑚/𝑠²  =1𝑚 =4𝑚...

 =1𝑚 =4𝑚 = 0.25 𝑚/𝑠 = 1.75 𝑚/𝑠 = 0.5 𝑚/𝑠² = 0.5 𝑚/𝑠² =1𝑚 = 𝟐. 𝟓 𝑚 = 0.25 𝑚/𝑠 = 1.25 𝑚/𝑠 = 0.5 𝑚/𝑠² = 0.5 𝑚/𝑠² =1𝑚 =4𝑚 = 0.25 𝑚/𝑠 = 1.75 𝑚/𝑠 = 0.5 𝑚/𝑠² = 0.5 𝑚/𝑠²  =1𝑚 =4𝑚 = 0.25 𝑚/𝑠 = 1.75 𝑚/𝑠 = 0.5 𝑚/𝑠² = 0.5 𝑚/𝑠² = Δx = x – x0 = 4 – 1 = 3 m = vavg = Δx / Δt = ( 3 m ) / ( 3 s ) = 1 m/s = aavg = Δv / Δt = ( v – v0 )/ Δt = ( 1.75 m/s – 0.25 m/s ) / ( 3 s ) = 0.5 m/s2 𝑚/𝑠 𝑚/𝑠 𝑚/𝑠 𝑚/𝑠 𝑎 = 0 + 𝑎𝑡 = 0 + 0𝑡 + ½ 𝑎 𝑡2 2 – 2 = 2𝑎( – 0 0) 𝑥0 = 1000 𝑚 𝑣0 = 0 𝑚/𝑠 𝑎 = −9.8 𝑚/𝑠2 𝑥 = 0𝑚 𝑥0 = 1000 𝑚 𝑣0 = 0 𝑚/𝑠 𝑎 = −9.8 𝑚/𝑠2 𝑎 = −9.8 𝑚/𝑠2 𝒗 =? 𝑥 = 0𝑚 𝑥0 = 1000 𝑚 𝑣 = 𝑣0 + a t 𝑣0 = 0 𝑚/𝑠 x = x0 + 𝑣0 t + ½ 𝑎 t2 𝑎 = −9.8 𝑚/𝑠2 𝒗𝟐 – 𝒗02 = 2 𝒂 ( x – x0 ) 𝑎 = −9.8 𝑚/𝑠2 𝒗 =? 𝑥 = 0𝑚 𝑥0 = 1000 𝑚 𝑣2 – 𝑣02 = 2 𝑎 ( – 0) 𝑣0 = 0 𝑚/𝑠 𝑣2 = 𝑣02 + 2 𝑎 ( – 0) 𝑎 = −9.8 𝑚/𝑠2 𝑎 = −9.8 𝑚/𝑠2 𝒗 =? 𝑥 = 0𝑚 𝑥0 = 1000 𝑚 𝑣2 – 𝑣02 = 2 𝑎 ( – 0) 𝑣0 = 0 𝑚/𝑠 𝑣2 = 𝑣02 + 2 𝑎 ( – 0) 𝑎 = −9.8 𝑚/𝑠2 𝑣2 = 19600 𝒗 = −𝟏𝟒𝟎 𝒎/𝒔 𝑎 = −9.8 𝑚/𝑠2 𝒗 =? 𝑥 = 0𝑚 𝑡 𝑡 𝑡  Δ / Δ𝑡 Δ / Δ𝑡 (𝑚) ( 𝑚/𝑠 ) 𝑎 ( 𝑚/𝑠 ) 𝑡(𝑠) 𝑡(𝑠) 𝑡(𝑠)  Δ / Δ𝑡 Δ / Δ𝑡 (𝑚) ( 𝑚/𝑠 ) 𝑎 ( 𝑚/𝑠 ) 𝑡(𝑠) 𝑡(𝑠) 𝑡(𝑠)  Δ / Δ𝑡 Δ / Δ𝑡 (𝑚) ( 𝑚/𝑠 ) 𝑎 ( 𝑚/𝑠 ) 𝑡(𝑠) 𝑡(𝑠) 𝑡(𝑠)  Δ / Δ𝑡 Δ / Δ𝑡 (𝑚) ( 𝑚/𝑠 ) 𝑎 ( 𝑚/𝑠 ) 𝑡(𝑠) 𝑡(𝑠) 𝑡(𝑠)  Δ / Δ𝑡 Δ / Δ𝑡 (𝑚) ( 𝑚/𝑠 ) 𝑎 ( 𝑚/𝑠 ) 𝑡(𝑠) 𝑡(𝑠) 𝑡(𝑠)  Δ / Δ𝑡 Δ / Δ𝑡 (𝑚) ( 𝑚/𝑠 ) 𝑎 ( 𝑚/𝑠 ) 𝑡(𝑠) 𝑡(𝑠) 𝑡(𝑠)  Δ / Δ𝑡 Δ / Δ𝑡 (𝑚) ( 𝑚/𝑠 ) 𝑎 ( 𝑚/𝑠 ) 100 𝑘𝑚 50 𝑘𝑚/ℎ 𝑡(𝑠) 𝑡(𝑠) 𝑡(𝑠) 2ℎ 2ℎ  Δ 𝑎𝑣𝑔 Δ𝑡 Δ / Δ𝑡 Δ / Δ𝑡 (𝑚) ( 𝑚/𝑠 ) 𝑎 ( 𝑚/𝑠 ) 100 𝑘𝑚 50 𝑘𝑚/ℎ 𝑡(𝑠) 𝑡(𝑠) 𝑡(𝑠) 2ℎ  Δ / Δ𝑡 ( 𝑚/𝑠 ) Δx = A1 + A2 + A3 + A4 𝑡(𝑠) Δt   a = –g a = –9.8 m/s2 ax = 0 ay = –g vx = v0x vy = v0y – g t x = x0 + v0x t y = y0 + v0y t – ½ g t2  θ sin(θ) = opp. / hyp. opp. = (hyp) sin(θ) v0y = v sin(θ) cos(θ) = adj. / hyp. adj. = (hyp) cos(θ) v0x = v cos(θ) tan(θ) = opp. / adj.  a = –g a = –9.8 m/s2 θ θ ax = 0 ay = –g vx = v0x vy = v0y – g t x = x0 + v0x t y = y0 + v0y t – ½ g t2  v0x = v cos(θ) v0y = v sin(θ)      Force I Newton‘s Laws of Motion Learning Objectives You will understand Newton’s 3 laws. You will be familiar with units of force. You’ll be able to solve 1D and 2D force problems. Force Definition Name Symbol Units Unit Abbreviation Mass m kilograms kg Position x meters m Velocity v meters per second m/s Acceleration a meters per second, per second m/s2 Newton‘s Laws Definition 1. Inertia An object‘s velocity will not change, unless acted on by an external force. 2. Force and acceleration The sum of forces on an object will be equal to its mass times its acceleration: 𝐹𝐹 = 𝑚𝑚 𝑎𝑎 3. Action-reaction For every force, there is a reaction force equal in magnitude and opposite in direction. Barrington Bramley, PD Force – Notation Vectors A Velocities Forces Force Working on an Apple 1. Inertia m1 v v = constant 2. Force and acceleration F=ma [ F ] = (kg) (m/s2) = Newtons, N m1 F 3. Action-reaction F1 on 2 F1 on 2 = -F2 on 1 m1 m2 F2 on 1 Equal and Opposite Forces M atruck = –m abug M m atruck abug Newton‘s 2nd Law Example, 1 Dimension Force in 1 Dimension Example You accelerate your 1,000 kg spaceship with only 1 N of force, starting from rest in space. How far has the spaceship traveled in 30 days?  Example x0 = 0 x=? F=1N v0 = 0 How far has the spaceship traveled in 30 days? x0 = 0 x=? F=1N v0 = 0 Kinematics Versus Dynamics Equations Kinematics Dynamics Equations of motion Acceleration Newton’s 2nd law x = x0 + v0 t + ½ a t2 a v = v0 + a t F=ma v2 - v02 = 2 a (x – x0) Newton‘s 2nd Law Example, 2 Dimensions Newton‘s 2nd Law Example You accelerate a 1,000 kg spaceship with 2 N, starting from rest in space. Now, a leak exerts a force of 1 N at an angle of 30 degrees from the ship’s direction of motion. 1. What is the spaceship’s velocity after 1 hour? 2. What is its speed? Force Problems in 2 Dimensions Example Fb = 1 N θ = 30o Fa = 2 N Force Problems in 2 Dimensions II 1. 2. 3. 4. 5. Choose a Draw a free Break all vectors Write Newton’s Solve for coordinate body diagram into perpendicular 2nd law for acceleration system components along each axis chosen axes Force in 2 Dimensions – 1st Step Choose a Coordinate System Equations +y +x Force in 2 Dimensions – 3rd Step Break All Vectors Equations +y Fb Fb θ Fb sin(θ) θ +x Fa Fb cos(θ) Force in 2 Dimensions – 3rd Step Break All Vectors Equations +y Fb θ Fb sin(θ) +x Fa Fb cos(θ) Force in 2 Dimensions – 3rd Step Break All Vectors Equations +y Fb Fb sin(θ) θ +x Fa Fb cos(θ) Force in 2 Dimensions – 3rd Step Break All Vectors Equations +y Fb sin(θ) +x Fa + Fb cos(θ) Force in 2 Dimensions – 4th Step Newton’s 2nd Law Equations F=ma +y ( Fx , Fy ) = m ( ax , ay ) x: Fx = m ax Fa + Fb cos(θ) = m ax y: Fy = m ay Fb sin(θ) = m ay Fb sin(θ) +x Fa + Fb cos(θ) Force in 2 Dimensions – 5th Step Solve for Acceleration Equations F=ma +y ( Fx , Fy ) = m ( ax , ay ) x: Fx = m ax Fa + Fb cos(θ) = m ax y: Fy = m ay Fb sin(θ) = m ay Fb sin(θ) +x ax = [ Fa + Fb cos(θ) ] / m Fa + Fb cos(θ) ay = [ Fb sin(θ) ] / m Newton‘s 2nd Law Example You accelerate a 1,000 kg spaceship with 2 N, starting from rest in space. Now, a leak exerts a force of 1 N at an angle of 30 degrees from the ship’s direction of motion. 1. What is the spaceship’s velocity after 1 hour? 2. What is its speed? Force in 2 Dimensions – 5th Step Solve for Acceleration Equations v=? +y v = v0 + a t vx = ax t = [ Fa + Fb cos(θ) ] t / m vy = ay t = [ Fb sin(θ) ] t / m Fb sin(θ) +x ax = [ Fa + Fb cos(θ) ] / m Fa + Fb cos(θ) ay = [ Fb sin(θ) ] / m Force in 2 Dimensions – 5th Step Solve for Acceleration Equations v=? +y v = v0 + a t vx = [ 2 + 1 cos(30o) ] 3600 / 1000 ~ 10.3 m/s vy = [ 1 sin(30o) ] 3600 / 1000 = 1.8 m/s Fb sin(θ) +x ax = [ Fa + Fb cos(θ) ] / m Fa + Fb cos(θ) ay = [ Fb sin(θ) ] / m Force in 2 Dimensions – 5th Step Solve for Acceleration Equations speed = ? +y c² = a2 + b2 Fb sin(θ) +x c Fa + Fb cos(θ) b a Force in 2 Dimensions – 5th Step Solve for Acceleration Equations speed = ? +y vx = 10.3 m/s vy = 1.8 m/s speed = |v| = 𝑣𝑣𝑥𝑥2 + 𝑣𝑣𝑦𝑦2 = 10.46 m/s Fb sin(θ) |v| = 10.46 m/s +x Fa + Fb cos(θ) vy = 1.8 m/s vx = 10.3 m/s Sine and Cosine Values for the Important Angles Exam Angle x, 0° or 30° 45° 60° 90° 180° degrees 360° Angle x, π π π π 0 or 2π π radians 6 4 3 2 1 2 3 sin(x) 0 1 0 2 2 2 3 2 1 cos(x) 1 0 -1 2 2 2 Newton‘s 3rd Law Example Newton‘s 3rd Law Example A force from the left (FT) pushes two touching objects FT (m1 and m2) as shown here. m1 If m1 = 10kg, m2 = 5kg, and FT = 15N, what is the force m2 of m1 on m2? Newton‘s 3rd Law Example Total system F = ma FT = ( m1 + m2 ) a a = FT / ( m1 + m2 ) = 15 / ( 10 + 5 ) = 1 m/s2 m1 m2 FT = 15 N F1 = m a F2 = m a m1 = 10 kg m2 = 5 kg FT + F2 on 1 = m1 a F1 on 2 = m2 a F2 on 1 = m1 a – FT F1 on 2 = ( 5 ) ( 1 ) F2 on 1 = 10 – 15 F1 on 2 = 5 N F2 on 1 = –5 N Learning Outcomes  Know Newton‘s 3 laws.  Understand the units of force.  Be able to solve simple force problems. Force II Important Forces Learning Objectives You will be familiar with different types of forces. You will be able to apply Newton's Laws to these forces. Important Forces Gravity Normal Tension Friction Air force resistance Gravity and the Normal Force Definitions Gravity Equations ! 𝑚𝑚1 𝑚𝑚2 𝐹𝐹 = 𝐺𝐺 𝑟𝑟 2 Barrington Bramley, PD Gravity Equations 𝑚𝑚1 𝑚𝑚2 Mount Everest 𝐹𝐹 = 𝐺𝐺 𝑟𝑟 2 𝑀𝑀𝐸𝐸 𝑚𝑚 RE 𝐹𝐹 = 𝐺𝐺 𝑅𝑅𝐸𝐸2 𝑀𝑀𝐸𝐸 𝑀𝑀𝐸𝐸 𝐹𝐹 = 𝐺𝐺 𝑚𝑚 𝑔𝑔 ≡ 𝐺𝐺 𝑅𝑅𝐸𝐸2 𝑅𝑅𝐸𝐸2 Ocean floor g Fg = m g (downwards) https://pixabay.com/en/globe-earth-world-transparency-1348777/ Normal Force Definition FN ainto surface = 0 Fg = m g Definition The Inclined Plane Setup Important Forces Fg FN Gravity Normal Tension Friction Air force resistance The Inclined Plane Definition +y FN +x θ θ Fg cos(θ) Fg sin(0) = 0 θ Fg Fg sin(θ) θ θ cos(0) = 1 The Inclined Plane Example The Inclined Plane and Gravity Example A block with weight of 10 kg slides down a smooth inclined plane, at an angle of 30º with the horizontal. How long will it take the block to slide 5 meters down the slope? How long will it take the block to slide 5 meters down the slope? +y FN +x 10 kg Fg cos(θ) Fg sin(θ) 5m θ θ = 30o How long will it take the block to slide 5 meters down the slope? +y FN +x 10 kg Fg cos(θ) Fg sin(θ) 5m θ θ = 30o Tension Definition Important Forces Fg FN Gravity Normal Tension Friction Air force resistance Tension Definition FT Tension force Fg Tension Definition Pulley FT FT Fg Fg Tension Example The Inclined Plane and Tension Example A 10 kg mass on a slope is connected to a rope, which goes over a pulley at the top and connects to a 15 kg mass. If the slope has angle 30º and the masses start from rest, how far will the 15 kg mass fall in 1 second? How far will the 15 kg mass fall in 1 second? FN FT FT M1 M2 Fg1 cos(θ) Fg1 sin(θ) +y Fg2 θ +x How far will the 15 kg mass fall in 1 second? FN FT FT M1 Fg1 cos(θ) Fg1 sin(θ) M2 +y Fg2 θ +x Friction Definition Important Forces Fg FN FT Gravity Normal Tension Friction Air force resistance Friction Definition Ff = μ F N Friction Definition Big μ Small μ FN Rough Smooth Ff v Rough Smooth Ff = μ F N Friction Equations Kinetic friction Static friction Ff = μk FN Ff ≤ μs FN FN Ff FN Ff v Friction Definition Kinetic friction Static friction μk < μ s Ff = μk FN Ff ≤ μs FN Kinetic friction is weaker than static FN friction. Ff FN Ff v Friction Example The Inclined Plane and Friction Example A block is at rest on a slope with coefficient of static friction 0.5, and coefficient of kinetic friction 0.3. To get the block to the threshold of sliding downwards, what angle must the ramp be lifted to? To get to the threshold of sliding downwards, what angle must the ramp be lifted to? +y +x FN Ff M Fg cos(θ) Fg sin(θ) θ To get to the threshold of sliding downwards, what angle must the ramp be lifted to? +y +x FN Ff M Fg cos(θ) Fg sin(θ) θ Air Resistance Definition Important Forces Fg FN FT Ff (Drag) Gravity Normal Tension Friction Air force resistance Air Resistance (Drag) Definition Air resistance Terminal velocity Fd ∝ v 2 Fd Fd Fd Velocity dependent Falling Fd = Fg object Increases until equal to gravity for falling objects Fg Fg Fg Learning Outcomes  Know a variety of Force types  Understand the limits and applications of these Forces  Be prepared to solve Force problems in 1D and 2D  𝑎𝑐 = 𝑣 2 /𝑟 𝐹𝑐 = 𝑚 𝑎𝑐 = 𝑚 𝑣 2 /𝑟  m1 x1 + m2 x2 +... “Weighted” positions Center of mass = = m1 + m2 +... Total mass  2 × 1030 kg 6 × 1024 kg 1.5 × 108 km 7 × 1022 kg 4 × 105 km   Optics I Reflection and Refraction Learning Objectives You can understand the difference between reflection and refraction. You will know the definition of the index of refraction. You will know and be able to use Snell’s law. You will be able to derive the conditions for total internal reflection. Optics in Context I. Reflection and refraction II. Mirrors and lenses III. Optical instruments Reflection and Refraction Definition Index of Refraction Definition Light travels slower in different Light media than vacuum. Vacuum The index of refraction: 𝑐𝑐 Air 𝑛𝑛 = 𝑣𝑣 A higher 𝑛𝑛 means a slower Water velocity. Reflection and Refraction Definition Vacuum Air Reflected Refraction: Transmitted Light entering a medium with a different 𝑛𝑛 and will bend. Reflection: A portion of the light will be reflected from the surface. Incident Reflection and Refraction Definition Vacuum Air Reflected Transmitted 𝜃𝜃𝑖𝑖 : angle of incidence 𝜃𝜃𝑟𝑟 𝜃𝜃𝑡𝑡 𝜃𝜃𝑟𝑟 : angle of reflection 𝜃𝜃𝑖𝑖 𝜃𝜃𝑡𝑡 : angle of transmission The amount of reflection decreases Incident with decreasing 𝜃𝜃𝑖𝑖. Reflection and Refraction Definition Vacuum Air Reflected Transmitted The incident angle equals the reflected angle. 𝜃𝜃𝑟𝑟 𝜃𝜃𝑖𝑖 𝜃𝜃𝑖𝑖 = 𝜃𝜃𝑟𝑟 Incident Snell’s Law Definition Snell’s Law Definition 𝑛𝑛1 𝑛𝑛2 𝑛𝑛1 sin 𝜃𝜃1 = 𝑛𝑛2 sin 𝜃𝜃2 𝜃𝜃2 Going into a denser medium, 𝜃𝜃1 𝜃𝜃2 is smaller. Going into a sparser medium, 𝜃𝜃2 is bigger. Snell’s Law Definition 𝑛𝑛1 𝑛𝑛2 𝑛𝑛1 sin 𝜃𝜃1 = 𝑛𝑛2 sin 𝜃𝜃2 𝜃𝜃2 Going into a denser medium, 𝜃𝜃1 𝜃𝜃2 is smaller. Going into a sparser medium, 𝜃𝜃2 is bigger. Total Internal Reflection Equations 𝑛𝑛2 𝑛𝑛1 sin 𝜃𝜃1 = 𝑛𝑛2 sin 𝜃𝜃2 𝑛𝑛1 sin 𝜃𝜃1 = 𝑛𝑛2 sin 𝜃𝜃1 = 𝑛𝑛1 𝜃𝜃2 < 90° 𝜃𝜃2 = 90° 𝜃𝜃2 > 90° 𝑛𝑛2 𝑛𝑛1 𝜃𝜃1 𝜃𝜃1 𝜃𝜃1 Total Internal Reflection Definition 𝑛𝑛1 must be greater than 𝑛𝑛2 for total internal reflection. 𝑛𝑛2 sin 𝜃𝜃critical = 𝑛𝑛1 This is only possible traveling from a more dense medium. 𝜃𝜃2 < 90° 𝜃𝜃2 = 90° 𝜃𝜃2 > 90° 𝑛𝑛2 𝑛𝑛1 𝜃𝜃1 𝜃𝜃1 𝜃𝜃1 Dispersion Definition Dispersion Definition The index of refraction can depend on 𝑛𝑛1 the wavelength of the light. 𝑛𝑛2 𝑛𝑛red < 𝑛𝑛blue By Snell’s law, different colors will diffract differently. The separation of light is dispersion. Dispersion A prism uses dispersion to separate frequencies. Learning Outcomes  You understand the difference between reflection and refraction.  You know the definition of the index of refraction.  You know and are able to use Snell’s law.  You are able to derive the conditions for total internal reflection. Optics II Mirrors and Lenses Learning Objectives You will know the basic properties of mirrors and lenses. You will be able to find the image created by a mirror or lens. You can identify an image’s location and type. Optics in Context I. Reflection and refraction II. Mirrors and lenses III. Optical instruments Mirrors Properties Mirror Properties Definition Mirror We consider a curved circular mirror. Mirror Properties Definition Concave Convex These can be concave or convex. (like a “cave”) Mirror Properties Definition 𝑟𝑟 Circular mirrors have a center of 𝐶𝐶 curvature 𝐶𝐶 and a radius 𝑟𝑟. Mirror Properties Definition Parallel rays reflect through 𝐶𝐶 𝐹𝐹 the focal point 𝐹𝐹. Mirror Properties Definition 𝑟𝑟 The focal point is located distance 𝑓𝑓 from the mirror. 𝐶𝐶 𝐹𝐹 𝑟𝑟 The focal point is also 𝑓𝑓 =. 2 𝑓𝑓 Mirror Properties Definition The light from distant objects is approximately parallel upon arrival 𝐶𝐶 𝐹𝐹 Mirror Properties Definition Finding images of objects: Each point emanates light in all directions. Each light ray appears to be coming directly to the eye. Mirror Properties Definition Finding images of objects: Each point emanates light in all directions. Each light ray appears to be coming directly to the eye. Tip of arrow looks to be sending light from here Mirrors Finding Images Finding Images 1. From the tip of the arrow, send a parallel beam and a beam through the focus. 2. The parallel beam will go through the focus, and the focal beam will become parallel. 3. The point of intersection will appear to be the new source of the light. 4. From the base of the arrow, the result is just a line at the base. 5. The resulting image is an upside-down arrow. Finding Images 1. From the tip of the arrow, send a parallel beam and a beam through the focus. 2. The parallel beam will go through the focus, and the focal beam will become parallel. 3. The point of intersection will appear to be the new source of the light. 4. From the base of the arrow, the result is just a line at the base. 5. The resulting image is an upside-down arrow. Finding Images 1. From the tip of the arrow, send a parallel beam and a beam through the focus. 2. The parallel beam will go through the focus, and the focal beam will become parallel. 3. The point of intersection will appear to be the new source of the light. 4. From the base of the arrow, the result is just a line at the base. 5. The resulting image is an upside-down arrow. Finding Images 1. From the tip of the arrow, send a parallel beam and a beam through the focus. 2. The parallel beam will go through the focus, and the focal beam will become parallel. 3. The point of intersection will appear to be the new source of the light. 4. From the base of the arrow, the result is just a line at the base. 5. The resulting image is an upside-down arrow. Finding Images 1. From the tip of the arrow, send a parallel beam and a beam through the focus. 2. The parallel beam will go through the focus, and the focal beam will become parallel. 3. The point of intersection will appear to be the new source of the light. 4. From the base of the arrow, the result is just a line at the base. 5. The resulting image is an upside-down arrow. Convex Mirrors Definition Rays towards the focal point leave parallel. Rays parallel reflect directly away from the focal point. The image is at the crossing of the extended rays. This image is virtual, it cannot be projected onto a screen. Convex Mirrors Definition Rays towards the focal point leave parallel. Rays parallel reflect directly away from the focal point. Image The image is at the crossing of the extended rays. This image is virtual, it cannot be projected onto a screen. Mirror Properties Definition 𝑜𝑜 We can quantify the locations of the Object Object Image 𝑓𝑓 Image Focus 1 1 1 𝑖𝑖 + = 𝑜𝑜 𝑖𝑖 𝑓𝑓 Mirror Properties 𝑜𝑜 Conventions: 𝑓𝑓: (+) for concave mirrors, (-) for convex Object 𝑜𝑜: always (+) 𝑓𝑓 Image 𝑖𝑖: (+) for real images in front of the mirror, (-) for virtual images behind the mirror 𝑖𝑖 Concave Mirrors – Magnification Definition 𝑖𝑖 𝑜𝑜 𝑚𝑚 = − 𝑜𝑜 Object Negative magnification means 𝑓𝑓 inverted images. Image Real images are inverted. 𝑖𝑖 Virtual images are upright. Concave Mirrors – Magnification Definition 𝑖𝑖 𝑚𝑚 = − 𝑜𝑜 𝑖𝑖 Negative magnification means inverted images. Real images are inverted. 𝑜𝑜 × 𝑚𝑚 Virtual images are upright. Magnification Triangle: Cover a variable to see its equation. Concave Mirrors – Magnification Definition 𝑖𝑖 ℎ𝑖𝑖 𝑚𝑚 = − = 𝑜𝑜 ℎ𝑜𝑜 Object The magnification is also the ratio ℎ𝑜𝑜 Image of the heights of the images. ℎ𝑖𝑖 Example: Original object is 1m tall, and the image appears 2m tall. ℎ𝑖𝑖 2 𝑚𝑚 = = =2 ℎ𝑜𝑜 1 Virtual images are upright Concave Mirror – Summary Location of object Image type Infinite distance At the focal point Outside focus Inverted, real On focus Image goes to infinity Inside focus Upright, virtual Concave Mirror – Summary Location of object Image type Infinite distance At the focal point Outside focus Inverted, real On focus Image goes to infinity Inside focus Upright, virtual Concave Mirror – Summary Location of object Image type Infinite distance At the focal point Outside focus Inverted, real On focus Image goes to infinity Inside focus Upright, virtual Concave Mirror – Summary Location of object Image type Infinite distance At the focal point Outside focus Inverted, real On focus Image goes to infinity Inside focus Upright, virtual Convex Mirror – Summary Location of object Image type Anywhere Virtual, upright Lenses Properties Lenses Definition Lenses follow the same logic as mirrors. They can be converging or diverging. Lenses Definition Diverging lens Converging lens Lenses follow the same logic as mirrors. They can be converging or diverging. Lenses Definition 𝐶𝐶 𝐹𝐹 Lenses follow the same logic as mirrors. Each side has a center of curvature and a focal point. Converging Lenses Definition We consider thin lenses. The same geometries apply with the focal point: The first ray begins parallel, 𝐶𝐶 𝐹𝐹 and passes through the focal point. The second ray goes through the focal point, then becomes parallel. Converging Lenses Definition We consider thin lenses. The same geometries apply with the focal point: The first ray begins parallel, and passes through the focal point. The second ray goes through the focal point, then becomes parallel. Diverging Lenses Definition The first ray begins parallel, then diverges from the focal point The second ray targets the far focal point, then diverges to become parallel The image is found by tracing the rays back Diverging Lenses Definition The first ray begins parallel, then diverges from the focal point The second ray targets the far focal point, then diverges to become parallel The image is found by tracing the rays back Converging Lens – Summary Location of object Image type Infinite distance At the focal point Outside focus Inverted, real On focus Image goes to infinity Inside focus Upright, virtual Converging Lens – Summary Location of object Image type Infinite distance At the focal point Outside focus Inverted, real On focus Image goes to infinity Inside focus Upright, virtual Converging Lens – Summary Location of object Image type Infinite distance At the focal point Outside focus Inverted, real On focus Image goes to infinity Inside focus Upright, virtual Converging Lens – Summary Location of object Image type Infinite distance At the focal point Original rays Outside focus Inverted, real On focus Image goes to infinity Inside focus Upright, virtual Diverging Lens – Summary Location of Object Image type Anywhere Virtual, upright Lens Properties Definition 𝑜𝑜 𝑖𝑖 We can quantify the locations of the Object Image Focus 1 1 1 + = 𝑜𝑜 𝑖𝑖 𝑓𝑓 𝑓𝑓 Lens Properties Definition 𝑜𝑜 𝑖𝑖 Conventions: 𝑓𝑓: (+) for converging lenses, (-) for diverging lenses 𝑜𝑜: always (+) 𝑖𝑖: (+) for real images, (-) for virtual images 𝑓𝑓 Learning Outcomes  You know the basic properties of mirrors and lenses.  You are able to find the image created by a mirror or lens.  You can identify an image’s location and type.   1 2 𝑟 𝑃= = 𝑓 𝑟 𝐶 𝐹 1 𝑃 = , Diopters 𝑚 𝑓  𝑛 𝑃total ~ 𝑛 𝐶 𝐹  1 1 𝑃total = 𝑃1 + 𝑃2 = + 𝑓1 𝑓2 𝐹1 𝐹2 𝑚total = 𝑚1 𝑚2 𝑓1 𝑓2  1 1 𝑃total = 𝑃1 + 𝑃2 = + 𝑓1 𝑓2 𝑚total = 𝑚1 𝑚2 𝑚1 𝑚2 𝐹1 𝐹2 𝑚1 = 2 𝑚2 = 3 𝑚 = 𝑚1 𝑚2 = 6     𝑛medium 1 𝑛lens 1 1 = −1 − 𝑓 𝑛medium 𝑟1 𝑟2 𝑟1 𝑛lens 𝑟2   𝜃2 𝑀𝜃 = 𝜃1 𝜃2 𝜃1 𝜃1  𝑐𝑚 𝜃 𝑀𝑜 = 𝜃25 𝑐𝑚        𝑟 𝐵 𝐼 𝐵∝ 𝑟 +𝐼 Force 𝐵 = Charge ∙ Velocity 𝑁 𝐵 = = Tesla, 𝑇 𝐴×𝑚  𝑟 𝐵 +𝐼 +𝐼  𝐼 𝐵 𝐵 +𝐼 𝐵 𝐵 𝐵  +𝐼 +𝐼  +𝐼 +𝐼 1 𝐵 ∝ 𝐼/𝑟  𝐿 𝑅=𝜌 𝐴 𝐼 2𝐼  𝑣 𝐹𝐵 = 𝑞 𝑣 × 𝐵 𝑞  𝐵 𝐹𝐵 = 𝑞 𝑣 𝐵 sin 𝜃 𝑣 𝜃 𝑞  𝐵 𝑣 𝑣 𝑞  𝐵 𝑣 𝑣 𝑞  𝐵 𝑣 𝑣 𝑞 𝐹𝐵 𝐹𝐵  𝐵 𝒗 𝑣 𝑣 𝑞 𝐹𝐵 𝐹𝐵  𝐸 𝐵 𝐹𝐸 𝐹 = 𝐹𝐸 + 𝐹𝐵 = 𝑞𝐸 + 𝑞 𝑣 𝐵 sin 𝜃 𝑣 𝑞 𝐹𝐵  𝑣2 𝑞 𝑣 𝐹=𝑚 𝑟 𝐹𝐵 𝐹𝐵 = 𝑞 𝑣 𝐵 𝑣2 𝑚 =𝑞𝑣𝐵 𝑟  𝑣2 𝑚 =𝑞𝑣𝐵 𝑣 𝑟 𝑞 𝑞𝐵𝑟 𝑣= 𝑚 𝐹𝐵 Distance 2 𝜋 𝑟 2 𝜋 𝑚 Time = = = Velocity 𝑣 𝐵 𝑞  𝑇 5.7 × 10−12 𝑘𝑔/𝐶  1 𝑚/𝑠 𝑞 𝑻 𝑹          𝜆 ~𝜆        𝐸 𝐵 𝜆  𝐸 𝑣 𝑣 𝐵  𝑚 𝑐 = 3 × 108 𝑠 𝜆 𝑓 𝑐 𝑐=𝜆𝑓  𝐸 𝐵  𝐸  𝐸  𝐸  𝐸  𝐸      Light III The Light Spectrum Learning Objectives You will understand the basic features of the electromagnetic spectrum. You will know the definition of a photon. Light in Context I. Wave phenomena II. Properties of radiation III. The light spectrum Electromagnetic Spectrum Definition Electromagnetic Spectrum Wavelength Radio Microwave Infrared Visible Ultraviolet X-ray 𝛾𝛾-ray (in m) 103 10-2 10-5 0.5 x 10-6 10-8 10-10 10-12 About the size of: Buildings Humans Honeybee Pinpoint Protozoans Molecules Atoms Atomic nuclei Frequency (in Hz) 104 108 1012 1015 1016 1018 1020 Electromagnetic Spectrum R. O. Y. G. B. I. V. Visible light Orange Green Indigo Infrared Red Yellow Blue Violet Ultraviolet 800 700 600 500 400 300 λ (nm) Radio waves Microwaves Infrared light Ultraviolet light X-rays 𝛾𝛾-rays Higher Frequencies Ultraviolet X-rays 𝛾𝛾-rays Invisible light of slightly Higher energy and small Considered radiation, higher energy than blue wavelengths, useful for and part of the dangerous probing matter to see cosmic rays filtered by the Causes sunburn, DNA internal structure atmosphere and magneto- damage, and cancer! sphere of the earth Can be ionizing! Ionizing! Lower Frequencies Infrared Microwaves Radio waves Lower frequency than Even lower energy Very long wavelengths, red light, and lower waves, useful for exciting useful for transmitting energy; infrared water molecules and signals as they are not as radiation is emitted by heating them up with easily absorbed any object with heat high amplitude microwaves (still at low energy) Photons Definition Photons Definition Photon In the early20th century, experiments indicated that light could only come in specific frequencies, having to be “quantized”. These individual ”packets” of light are called photons. Photons Definition A photon has a discrete amount of energy, given by its frequency: 𝐸𝐸 = ℎ 𝑓𝑓 Planck’s constant: ℎ = 6.6 × 10−34 𝐽𝐽 × 𝑠𝑠 Photons Definition Blue photons have… …small wavelengths. …high energy. Red photons have… …large wavelengths. …low energy. 𝐸𝐸 = ℎ 𝑓𝑓 Photons Equations The photon energy equation is often expressed in angular units: ℎ 𝐸𝐸 = 2 𝜋𝜋 𝑓𝑓 = ℏ 𝜔𝜔 2 𝜋𝜋 Reduced Planck’s constant: ℎ ℏ≡ 2 𝜋𝜋 Angular frequency in radians: 𝜔𝜔 ≡ 2 𝜋𝜋 𝑓𝑓 Photons Equations A photon’s wavelength can also be used to find its energy from the wave velocity equation 𝑣𝑣 = 𝑓𝑓 𝜆𝜆, since 𝑓𝑓 = 𝑐𝑐/𝜆𝜆 ℎ 𝑐𝑐 𝐸𝐸 = ℎ 𝑓𝑓 = 𝜆𝜆 Learning Outcomes  You understand the basic features of the electromagnetic spectrum.  You know the definition of a photon.