Untitled

Refraction at Convex Spherical Surface

• Considers a convex spherical refracting surface.
• µ₂ represents the refractive index of medium 2.
• μ₁ represents the refractive index of medium 1.
• P refers to the pole, and C stands for the center of curvature.
• PC is the principal axis of the convex surface.
• The object is in the rarer medium and the image is real.
• Begins by drawing AN perpendicular and defines angles α, β, and γ in respective triangles.
• ∠AOP = α, ∠AIP = β, and ∠ACP = γ.
• In triangle AOC, i = α + γ, based on the exterior angle property; this is equation (i).
• For small angles α, β, and γ, they can be replaced by their tangents.
• Equation (i) is rewritten as i = tan α + tan γ, denoted as equation (ii).
• Uses right-angled triangles ANO and ANC to express tan α as AN/NO and tan γ as AN/NC.
• Substituting in equation (ii), it becomes i = AN/NO + AN/NC, labeled as equation (iii).
• Considers NO ≈ PO and NC ≈ PC,
• Equation (iii) simplifies to i = AN/PO + AN/PC, now equation (iv).
• From triangle ACI, γ = r + β due to the exterior angle property, rearranged as r = γ - β.
• For small angles β and γ, r = tan γ - tan β, denoted as equation (v).
• From right-angled triangles ANC and ANI, tan γ = AN/NC = AN/PC and tan β = AN/NI = AN/PI.
• Substituting in equation (v), results in r = AN/PC - AN/PI, which is equation (vi).
• Snell's law is given by µ₁ sin i = µ₂ sin r.
• For small angles i and r, the equation simplifies to μ₁i = μ₂r.
• Substitutes the values of i and r from equations (iv) and (vi) and results in μ₁(AN/PO + AN/PC) = μ₂(AN/PC - AN/PI).
• Which simplifies to μ₁/PO + μ₁/PC = μ₂/PC - μ₂/PI.
• Final result: μ₁/PO + μ₂/PI = (μ₂ - μ₁)/PC
• Applying Cartesian sign conventions: PO = -u (object distance), PI = +v (image distance), and PC = +R (radius of curvature).
• The equation becomes -μ₁/u + μ₂/v = (μ₂ - μ₁)/R
• Alternative form: μ₂/v - μ₁/u = (μ₂ - μ₁)/R

Magnetic Field at a Point on the Axis of a loop

• Focuses a circular loop with radius 'a', centered at O, carrying current I.
• Defines point P on the loop's axis at distance OP = x from the center O with AB = dl representing a small current element.
• Angle ∠BCP (or ∠ACP) is 90°.
• Biot-Savart law states the magnetic field due to current element AB at point P as dB = (µ₀ / 4π) * (Idl x r) / r³.
• Since the angle between dl and r is 90°, the magnitude of dB is given by dB = (µ₀ / 4π) * (Idl / r²); this is equation (i).
• Another element A'B' = dl is located opposite AB and dB' is the Mag Field due to the current element A'B'.
• If ∠OPC = ∠OPC' = θ, then ∠ZPL = ∠Z'PM = θ.
• The cos θ components cancel out upon resolving, leaving only the sin θ components to add.
• Equation for B becomes B = ∮ dB sin θ = ∮ (µ₀ / 4π) * (Idl / r²) * sin θ.
• Which simplifies to B = (µ₀ / 4π) * (I / r²) * sin θ ∮ dl.
• As the integral of dl, denoted as § dl = 2πα
• Therefore, B = (µ₀ / 4π) * (I / r²) * sin θ * (2πα); B = (µ₀ / 4π) * (I / r²) * sin θ * (2πα); this is equal to µ₀ 2πIa²/4πr³.
• B = (µ₀ 2πIa²) / 4π(a² + x²)³/²
• Also for 'N' turns B = (µ₀ 2πNIa²) / 4π(a² + x²)³/²

Force between Two Infinitely Long Parallel Current Carrying Conductors

• Considers two infinitely long conductors X1Y1 and X2Y2 parallel with a distance r, carrying currents I1 and I2 in the same direction.
• The magnetic fields of wire 1 & 2 are B1 = (µ₀ / 4π) * (2I₁ / r) and B2 = (µ₀ / 4π) * (2I₂ / r).
• Wire 1 experiences a magnetic force due to field of wire 2 and vice versa.
• L is the length of wires on which F is calculated, thus force F1,2 on wire 1 due to wire 2's field is F1,2 = I1(L × B2). This equates to F1,2 = (µ₀ / 4π) * (2I1I2 / r) * L (-î).
• Force per unit length has been expressed from F1.2
• Which is F1.2/ L = µ₀ 2I1I2/4π r (-î). (i) Also F2,1 = I2(L x B1) = µ₀ 2I1I2/4π r x I2 x L (î) (Force on wire 2 due to magnetic field of wire 1) or F2.1/ L = µ₀ 2I1I2/4π r (î)
• Force per unit length has been expressed from F2.1
• The force per unit length on both wires is equal and opposite.

AC Through LCR Series Circuit

• Considers an LCR series circuit with instantaneous e.m.f E and current I.
• VL, VC, and VR are instantaneous voltages across inductor L, capacitor C, and resistor R, respectively.
• VL = IXL, VC = IXC, and VR = IR
• XL = ωL and XC = 1/ωC are reactances of inductor and capacitor, respectively, with ω as the angular frequency.
• OE = √(OA2 + AE2); OE = √(OA2 + OD2) (Pythagoras)
• Total voltage E is then E = √VR² + (VL - VC)²
• Substituting VR, VL and VC, E = √(IR)² + (IXL - IXC)²; E = I√R² + (XL - XC)²
• E = I√R² + (XL- XC)² simplifies to I = E/√R² + (XL - XC)² (i)
• Let Z = E/I (ii)
• Thus from equations (i) and (ii), Z = √R² + (XL - XC)²; Z = √R² + (@L - 1/ωC)²
• This gives the impedance of LCR-circuit; Z = √R² + (@L - 1/ωC)² (iii)
• Also phase angle Ø From right angled△ OAE; tan $ = AE/OA; tan $ = VL-VC/VR ; tan $ = IXL - IXC/IR
• The final results are tan $ = XL - XC/R and tan $ = @L– 𝟏/ωC/R which are derived to tan $ = XL - XC/R and tan $ = @L– 𝟏/ωC/R

Lens Maker's Formula

• Considering a convex lens with a refracting surface with μ₂ as the refractive index of the outer medium and μ₁ as the refractive index of the lens.
• Suppose 0 is a point object placed on the principal axis of the lens.
• The surface XP₁Y forms the real image I₁ (assuming that material of the lens extends beyond the face XP₁Y as such).
• It can be obtained* that μ₁/P10 + μ₂ /P1I1 = μ₂-μ₁/P1C1 (i)
• Since the lens is thin, the point P₁ lies very close to the optical center C of the lens.
• Therefore, we may write P₁O ≈ CO; P₁I₁ ≈ CI₁ and P₁C₁ ≈ CC1
• Equation becomes: μ₁/CO + μ₂ /CI₁ = μ₂-μ₁/CC1 (ii)
• The image formed by first refraction act as virtual object for 2nd surface refraction
• μ₂ /P2I1 + μ₁ /P2I = μ₁ - μ₂ /P2C2
• Substituting that P211 ≈ CI1, P2I ≈ CI and P2C2 ≈ CC2
• We derive at the equation μ₂ /CI₁ + μ₁ /CI = μ₁ - μ₂ /CC2 (iv)
• Adding the eq (ii) and (iv),

μ₁ /CO + μ₂ /CI₁ + μ₂ /CI₁ + μ₁ /CI = μ₂-μ₁ /CC1 + μ₁ - μ₂ /CC2 (v) or μ₁ /CO + μ₁ /CI = (μ₂ – μ₁)(𝟏/CC1 - 𝟏/CC2)

• Applying the new Cartesian sign conventions: CO = -u (object distance), CI = +v (Final image distance), CC₁ = +R₁ and CC₂ = -R₂ (Radii of curvature)
• From (v) derive at: μ₁/-u + μ₁ /+v = (μ₂ – μ₁)(𝟏/+R1 - 𝟏/-R2); this equates to: μ₁/-u + μ₁ /+v =(μ₂ – μ₁)(𝟏/R1 + 𝟏/R2).
• Dividing both sides of the above equation by µ₁; since μ₂/μ₁ = μ, we have 𝟏/v - 𝟏/u = ((μ-1)(𝟏/R1 - 𝟏/R2) (vi)
• Also if u = CF₁ = -f₁ (focal length ), then v = ∞
• Setting the above condition in the equation (vi), we have 1/-f₁ =((μ - 1)(𝟏/R1 + 𝟏/-R2)
• or: 1/f = (μ -1)((𝟏/R1 - 𝟏/R2)

Refraction through a Prism

• KTS = δ is the angle of deviation.
• Based on the law that ∠TQO = ∠NQP = i and ∠RQO = r1, we have ∠TQR = i-r₁.
• Also, <TRO = ∠NSE = e and ∠QRO = r2. Therefore, ∠TRQ = e - r2
• In triangle TQR, by exterior angle property: δ = ∠TQR + ∠TRQ. This can be rearranged to: δ = (i + e) – (r1 + r2) … (i)
• In triangle QRO, the sum of the angles is 180°.
• Therefore the equation (ii) r1 + r2 + ∠QOR = 180°
• In quadrilateral AQOR, with angle A + ∠QOR = 180 (iii)
• From the prior equations (ii) and (iii), r1 + r2 = A (iv)
• Using equation (iv), substitute for (r1 + r2): δ = (i + e) - A (v)
• δ = δm when the equation is in minimum deviation position with e = i and r₂ = r₁ = r = A/2 (say)
• Setting: δ = δm and e = i using equation (v); A+δm =i+iori = (A+δm)/2.
• The refractive index of the material (ªµg or simply µ) of the prism is given by µ = sini/ sinr or µ = sin(A + δm)/2 / Sin A/2

Bohr's Theory of Hydrogen Atom

• In a hydrogen atom, an electron with charge -e revolves around a nucleus with charge +e in a circular orbit.
• The electrostatic force of attraction between the nucleus and electron: Fₑ = 1/4πε₀ * e²/r² (i)
• The centripetal force, with mass 'm' and velocity 'v': F꜀ = mv²/r (ii)
• Equating the electrostatic and centripetal forces: mv²/r = 1/4πε₀ * e²/r² (iii)
• Bohr's quantization condition states angular momentum is quantized; mvr = nh/2π (iv)
• Energy of the electron in nth orbit of a hydrogen-like atom: En = −(1/4πε₀)² * 2π²Z²me⁴ / n²h²
• Combining all equations, v = (1/4πε₀) * (2πe² / nh) (vi)
• Ek = /4πε₀ * (e²/2r)
• Total energy revolving round the nucleus becomes E = Ek + Ep; E = -e²/8πε₀r or En = −(1/4πε₀)² * 2π²me⁴ / n²h²

Mass Defect

• The mass defect is the difference between the sum of the masses of the nucleons constituting a nucleus (Z protons and A-Z neutrons) and the actual mass of the nucleus, denoted as Δm.
• Mass of the nucleons is then Zmp + (A – Z)mn with : mp = mass of proton and mn = mass of neutron
• Relating mass to a nucleus using If mN(zXA) = mass of the nucleus of the atom zXA, we calculate the mass defect with : Δm = [Zmp + (A-Z)mn] - mn(zXA) (i)
• Expressing the prior formula, adding and subtracting the mass of Z electrons i.e. Zme on the R.H.S. of equation (i)
• the mass defect (Δm) = [Zm(₁H¹) + (A – Z)mn] – m(zXA) (ii)

Electric Field on Axial Line of an Electric Dipole

• Considers a electric dipole (consisting of charges -q and +q), separated by a distance 2a
• The electric field E at point P due to the dipole will be E = EA + EB- Where EA = the resultant of the electric fields EA (due to -q at point A) and EB (due to +q at the B)
• Electric field equates to:: E = ((EB) – (EA) ) (1); E= q/4πε₀(1/(r−a)² - 1/(+a)²) = 1/4πε₀ (q x 4ra/(r² – a²)²) where P = q(2a)(electric dipole moment)
• In vector form: Ē =2Pr/4 πε₀(r² – a²)² (i)
• With dipole of small length: Ē =(1/4 πε₀)(2P/r³)

Electric Dipole in Uniform Electric Field

• Considers an electric dipole consisting of charges -q and +q and of length 2a placed in a uniform electric field E making an angle 0 with the direction of the field.
• Fnet = F+q+F-q -q = 0 and E = qE(2asin 0) = q(2a) Esin 0
• With electric dipole moment noted by P; the prior formula is: τ = pEsin 0 in vector form is = PxE

Electric Field due to Charged Spherical Shell

• Assumes a thin spherical shell, radius R and centre O, with +q as the charge on the spherical shell.
• The Gaussian surface will be a sphere.
• Let E represents the electric point P
• The flux through the area in the enclosed area is given by: dΦ = E.dS- With constant given as dΦ = Eds
• Calculate overall electric flux sdΦ = Sd E dS with resulting area on the spherical of 4tr².
• In the form of equations derived from Guass therom - E = q / ε₀ /4πR² or 1/4περ x q/ r²

Electric potential at any point due to an electric dipole

• AB is indicated as a dipole w. charge -q -p/w charge +q. Charge is calculated at varying differences from Point P.
• Therefore, Net potential at any given point of the dipole is calc. using V = V1 + V2
• V = q/4πε₀ [(1/PB)-(1/PA)] = q /4πε₀ {q x 1/(r+acos0) -/(1/r- acos0)} V= P x Cos(a)/ 4π e o(r - a *Cos^ 2( Theta)). = Equation 2 The results given are for distance away given a smaller form.

Potential Energy of an Electric Dipole, When Placed in an Electric Field

• Let the Dipole be held among a direction making an angle 0 with the direction of an external uniform electric field E. Torque has acted.
• Therefore, rotational energy is given by: PE(COS thera 1. + cos thera 2.).
• Therefore PE is in a direction where it may be related when found in external fields.

Relation between Drift Velocity and External Fields

• With average velocity the equation the velocity and average velocity are given by the relations Ud = Average velocity/ time of applied field.
• Therefore , combining the length current and cross sections of two areas .
• We get that the Velocity due to the drifting is = ne E * time/mass

Capacitance

• The Die electric will be found in all areas where the capacitance is needed. V.V
• By introducing the die electric constant to the area , we derive C = E0 *A / distance

Heatstone Bridge

• Bridge is designed to contain a balance where (VA -VB = Voltage)
• In a bridge the equation where if the state is balance: :VA-VB = IPR .
• Dividing this by various equations and canceling all areas,
• We get the constant relationship found to be that voltage / all the other areas in the equation,
• The areas are all equal to 1 in these derivations.

Mutual inductance of Two Solenoids

• Mutal interactions are set to occur where I-1 and I - 2 are constant with given equations
• It is with coil and solidoids relationship
• There fore , calculating total flux: Mu-1 = 1 with given formula on coil and Mu 2. = 1
• We find that that they have similar value with ( Mu * Nu =Area where Solenoids ( A -12/1 and distance.)

AC Equations

• Current is known as Isin (WT) with all areas
• Therefore: dQ is set in relation to the potential Gradient. - or dQ Idt
• The charge that moves through the are is then T / Time- 2 where the AC occurs.
• Therfore , total Integral is above eqq with Time = 0.
• This derives all forms where Root- Meain Quare is equal or virtually in an anomailous location.

Magnification and Ratio ( concaves )

• Ratio, that is shown - is determined with by = Image/ Objet
• Where we can detereming how how the ration and total over is 2
• It is with equal values that can measure to find the relationship- The ratios are found for areas with in concaves surfaces,