Input characteristics show input current vs input voltage

Base current IB vs base-emitter voltage VBE

Keep collector-emitter voltage VCE constant

Energy of emitted photon = band gap energy E = hc/λ = 1.9 eV

Given hc = 1240 eV·nm

λ = 1240/1.9 = 652.6 nm = 6.5×10⁻⁷ m

At polarizing angle, reflected and refracted rays are perpendicular

ip + r = 90° where ip = 57°

Therefore r = 33°

Diffraction occurs when wave encounters obstacle

Significant diffraction when obstacle size ≈ wavelength

Too large or small obstacles don't show clear diffraction

For normal vision, object at 25 cm forms image at 75 cm

Use lens formula 1/f = 1/v - 1/u

1/f = 1/-75 + 1/25 = 2/75, so f = 37.5 cm

Use mirror formula 1/f = 1/v - 1/u and m = -v/u = 3

f = R/2 = 15 cm, substitute m = -3

Solve to get u = 20 cm

Refractive index n = c₀/v where c₀ is speed in vacuum

v = 1/√(με) and c₀ = 1/√(μ₀ε₀)

n = √(με/μ₀ε₀)

Calculate change in base current: ΔIB = ΔIE - ΔIC

β = ΔIC/ΔIB = 0.49/(0.50-0.49) = 49

Verify using IE = IC + IB relationship

Analyze function of each component

Determine logical sequence of operations

Rectification first, then smoothing, then regulation

For minimum deviation: i = e and r₁ = r₂

In equilateral prism, A = 60° = r₁ + r₂

Calculate i using n = sin(i)/sin(r) = √2

Use formula for angular width: θ = 2λ/a

Convert units: λ = 600×10⁻⁹m, a = 0.2×10⁻³m

Calculate θ = 2(600×10⁻⁹)/(0.2×10⁻³) = 6×10⁻³ rad

Use Fresnel distance formula: D = a²/λ

Convert units to meters

Calculate: D = (4×10⁻³)²/(400×10⁻⁹) = 40m

Consider coherence requirement

Different wavelengths have different phases

No fixed phase relationship between different colors

For erect and enlarged image, object must be between F and optical center

Distance must be less than focal length (20 cm)

Choose 15 cm as it satisfies conditions

Consider refractive index variation with wavelength

Shorter wavelength has higher refractive index

Violet light shows maximum apparent shift

Consider device characteristics

Zener operates in reverse breakdown

Maintains constant voltage in breakdown

Use Doppler effect formula: f' = f[(v±vo)/(v∓vs)] where v is sound speed

For approaching source: 500 = f(350/[350-50]), solve for f

For receding source: f' = 428.6(350/[350+50]) = 375 Hz

Consider frequency remains constant

λ = v/f where v decreases in denser medium

Since f is constant, λ must decrease

Use Malus's law: I = I₀cos²θ

Given I = I₀/2, solve for θ

cos²θ = 1/2 gives θ = 45°

Consider transistor structure

Emitter supplies majority carriers

Must be heavily doped for sufficient carriers

Compare energy gaps

Semiconductors have smaller gaps than metals

Germanium has gap of 0.72 eV

Consider wavelength dependence of fringe width

β = λD/d where λ_blue < λ_yellow

Smaller wavelength gives smaller fringe width

Apply Rayleigh scattering law

Intensity I ∝ 1/λ⁴

This explains why sky is blue (shorter wavelengths scattered more)

Use mirror formula: 1/f = 1/u + 1/v

Calculate f = R/2 = 7.5 cm, then find v

Calculate magnification: m = -v/u

Consider light ray direction

Distances measured in direction of incident ray are positive

Distances opposite to incident ray are negative

Calculate I_C using Ohm's law: I_C = V_CE/R_C

Use β = I_C/I_B to find I_B

Check units and magnitudes

Use n_i² = np for semiconductors

Given initial n_i = 3×10¹⁶, p = 4.5×10²²

Calculate n = n_i²/p

Substitute A = 1, B = 0

Calculate Ā = 0

Add: 0 + 0 = 0 = B

Understand CE configuration

Note biasing requirements

Identify input junction

Use formula for fringe width: β = λD/d

Calculate separation for both wavelengths

Find difference for 4th order

Write equations for position of bright bands: nλ₁ = (n+1)λ₂

Substitute values: n(780) = (n+1)(520)

Solve for n: n = 2

Remember diffraction angle depends on wavelength

Blue light has shorter wavelength than yellow

Use formula θ = λ/a to compare

Use diffraction formula for single slit

Angular width of principal maximum = λ/a

Note this is for first minimum

Use lens formula: 1/f = 1/v - 1/u

Put u = -25 cm, v = 75 cm

Calculate f = (u×v)/(u+v) = 18.75 cm

Use mirror formula: 1/f = 1/v + 1/u

Put f = -10 cm, u = -20 cm

Solve for v: v = -20 cm

Recall mirror formula: 1/f = 1/v + 1/u

Put u = f in the formula

Note that when object is at F, no image is formed

At minimum deviation, i = e and i₁ = i₂

A = 2i - δm where A is apex angle

Calculate i = (A + δm)/2 = 49°

Radio waves bend due to varying refractive index

Similar to light bending in heated air

Both involve gradual change in refractive index

Conductivity σ = ne²τ/m where n is carrier density and τ is relaxation time

Temperature increase causes: 1) More electrons excited to conduction band (n increases exponentially) 2) More collisions (τ decreases)

Net effect is increase in conductivity as n increase dominates τ decrease

Use I = I₀cos²(πx/λ)

For λ/3, phase difference = 2π/3

Calculate I = K/4

In semiconductors, covalent bonds hold electrons between atoms. When electron leaves bond, creates vacancy.

This vacancy (hole) can move as other electrons fill it, leaving new hole. Acts like positive charge.

Hole is not a physical particle but absence of electron in covalent bond.

Introduction of mica creates path difference: Δx = t(μ-1) where t is thickness and μ is refractive index. Calculate Δx = 1×10⁻⁶(1.5-1) = 0.5×10⁻⁶ m.

Shift in fringe position is given by: y = DΔx/d where D is screen distance and d is slit separation.

Calculate: y = (1.2×0.5×10⁻⁶)/(2.4×10⁻³) = 0.25×10⁻³ m = 0.25 mm

Start with angular resolution formula: θ = d/D where θ is minimum angle (in radians), d is separation, D is distance. Convert 1 minute to radians: θ = (1/60)×(π/180) radians.

Use the formula: 3.14/D = (1/60)×(π/180). Here 3.14 m is the separation (d) and D is the distance we need to find.

Solve for D: D = 3.14×60×180/π = 10.8 km. This is maximum distance for distinct vision.

For unpolarized light through first polaroid: I₁ = I₀/2 = 128/2 = 64 Wm⁻².

For second polaroid: I₂ = I₁cos²θ = 64cos²45° = 32 Wm⁻².

For third polaroid: I₃ = I₂cos²45° = 32cos²45° = 16 Wm⁻². Each polaroid reduces intensity according to Malus's law.

At minimum deviation: i = e and r₁ = r₂ = A/2. Also, sin(i) = n sin(r) where n = cot(A/2).

Use relation: d = 2i - A where d is deviation angle. At minimum deviation, i = A/2 + d/2.

Solve to get d = 180° - 2A. The given n = cot(A/2) leads to this specific result.

Use lens formula: 1/f = 1/u + 1/v. As object moves, u changes uniformly but relationship with v is not linear.

When object is at focus, image is at infinity. As object approaches focus, v changes non-linearly due to inverse relationship.

Image moves away with increasing speed (non-uniform acceleration) due to non-linear nature of lens equation.

First calculate total energy received per second: Power = Intensity × Area = 20 W/cm² × (25×15) cm² = 7500 W

For perfectly reflecting surface, momentum change is 2E/c (twice incident momentum due to reflection). E = Power × time = 7500 J

Calculate: p = 2E/c = 2×7500/3×10⁸ = 5×10⁻⁵ kg m/s

Let's analyze each statement. Statement 1: Correct - forward bias reduces depletion width by pushing carriers towards junction.

Statement 2: Correct - reverse bias increases barrier width and potential. Statement 3: Correct - photodiodes operate in reverse bias for better photon detection.

Statement 4: Incorrect - LEDs are heavily (not lightly) doped to increase recombination probability and light emission efficiency.

Let's analyze each statement. In single slit diffraction, the angular separation between successive minima increases, making fringes unequal in width. Statement (i) is correct.

The intensity pattern follows a sinc² function, meaning central maximum has highest intensity and secondary maxima have decreasing intensities. Statement (iv) is incorrect.

Light energy is indeed conserved - the total energy in the pattern equals incident energy, just redistributed. Statement (iii) is correct. Statement (ii) about equal width is incorrect.

Let's use the condition for first diffraction minimum: asin θ = λ. Here θ = 30° and λ = 6500Å.

Substitute the values: a×sin(30°) = 6500Å. Remember to convert Å to appropriate units (1Å = 10⁻¹⁰m).

Solving for 'a': a = 6500×10⁻¹⁰/0.5 = 1.3×10⁻⁶m = 1.3 micron

First, find the position of the initial image: Since object is at infinity, first image forms at focal point of convex lens (30cm). Image size is 2cm. This image acts as object for concave lens.

For concave lens, u = 30-26 = 4cm (distance from concave lens to first image). Using lens formula 1/f = 1/v - 1/u with f = -20cm.

Calculate new image size using magnification formula m = v/u. The negative sign of focal length and proper substitution gives final image size of 2.5cm.

In air, the focal length depends on (µg-1) where µg = 3/2. Let's call this f₁.

In water, we need to use relative refractive index µg/µw = (3/2)/(4/3) = 9/8. Now focal length depends on (µg/µw-1) = 1/8.

The ratio fw:fa = (µg-1):(µg/µw-1) = (1/2):(1/8) = 4:1. This shows the focal length increases by a factor of 4 when the lens is placed in water.

When light enters a new medium, let's consider what actually changes. The frequency of light is determined by the source and remains constant across media.

The speed of light changes in different media due to different refractive indices. This change in speed causes the wavelength to change (since v = fλ), which results in bending of light at the interface.

The elasticity of the medium affects the speed, but it's not the direct cause. Amplitude changes don't cause bending. The fundamental cause is the change in wave speed.

Semiconductors have resistivity between conductors (10⁻⁶Ω cm) and insulators (10¹⁰Ω cm)

At room temperature, typical range is 10⁻³-10⁻⁶Ω cm

This allows useful electronic properties - not too conducting, not too insulating!

Energy gap is minimum energy needed to excite electron from valence to conduction band

For germanium (Ge), this gap is characteristic of semiconductor materials - smaller than insulators but larger than conductors

At 0K, Eg = 0.71 eV for Ge. Silicon has 1.1 eV, for comparison

Remember dispersion - different colors have different refractive indices

Focal length f = R/(μ-1). Refractive index μ is least for red light

Therefore, red light has maximum focal length. This is why red forms outer ring in chromatic aberration!

CD surface has microscopic tracks forming a diffraction grating. When white light hits these tracks...

Different wavelengths (colors) are diffracted at different angles according to d(sinθ) = nλ

This creates a spectrum of colors - similar to how a prism splits white light but through diffraction!

Consider light spreading out from a point source in all directions. Area of wavefront increases with square of distance.

Since energy must be conserved and spreads over increasing area, intensity must decrease as 1/r²

This is inverse square law - similar to gravitational and electric fields!

In Fraunhofer diffraction, light waves from different parts of slit interfere.

Central maximum is brightest because all waves arrive in phase. Secondary maxima occur with decreasing intensity.

Pattern shows central bright band with alternating dark and bright bands of decreasing intensity on both sides.

Fringe width β = λD/d where λ is wavelength, D is screen distance, d is slit separation.

Red light has longer wavelength than violet (λred ≈ 2λviolet). Other factors (D,d) remain same.

Therefore, βred/βviolet = λred/λviolet ≈ 2. Fringe width for red is double!

Use de Broglie wavelength λ=h/mv for both cases

Compare fringe widths β=λD/d

Since meλp so β₁>β₂

Analyze image formation by objective lens (real, inverted image)

Study image formation by eyepiece (virtual image of objective's image)

Final image remains inverted but is virtual and magnified

Use lens formula 1/v - 1/u = 1/f and magnification m = -v/u. Also given u + v = x

From m = -v/u, we get v = -mu. Substitute into u + v = x: u - mu = x

Solve for u, then use lens formula to get f = mx/(m+1)²

For concave lens, power is negative. Use lens maker's formula: 1/f = (μ-1)(1/R₁ + 1/R₂). Here R₁ = -R₂ = R

Power = -4.5 D means f = -22.22 cm. For equi-concave, both radii are equal in magnitude but opposite in sign.

Solve: -4.5 = (0.6)(2/R), giving R = -26.6 cm. Check sign convention!

In equilateral prism, all angles are 60°. Given i₁ = i₂ = (3/4)×60° = 45°

Total deviation D = i₁ + i₂ - A, where A is prism angle (60°)

D = 45° + 45° - 60° = 30°. This is minimum deviation condition as i₁ = i₂!

Analyze external circuit current

Electrons flow in wire

No other carriers in metal

Input frequency = 50 Hz

Each diode conducts for half cycle

Conduction frequency = 50 times/second

Use I=4I₀cos²(πx/β)

Substitute x=β/3

Calculate final intensity

First polaroid: I₁=I/2

Second polaroid: I₂=I₁cos²θ

I/4=I/2cos²θ → θ=45°

Critical angle sin⁻¹(1/μ)

μ varies with wavelength

Red has least μ, hence max critical angle

Analyze each optical element

Convex mirror: All reflected rays diverge

Image always behind mirror (virtual)

According to Huygens principle, all wavelets travel with same speed

Primary wavefront is envelope of secondary wavelets

Both must have same velocity for wave propagation

Use lens formula and differentiate

When image speed equals object speed, v=u

Solve for object distance: u=2f=20cm

Use lens maker's formula with μ₁/μ₂ ratio

Calculate new focal length with water

Percentage change = [(fl-fa)/fa]×100 = 300%

Consider forward bias voltage drop 0.7V

Calculate effective voltage: 5.7-0.7=5V

Current=5V/5kΩ=1mA

Understand formation of depletion region

Consider recombination of charge carriers

Region is depleted of mobile charges

Use μ=sin(A+dm)/2/sin(A/2) where A=60° and dm=60°

Calculate μ=sin60°/sin30°=√3

Find speed using c=c₀/μ=3×10⁸/√3=√3×10⁸ ms⁻¹

Compare band structures of different materials

Identify that conductors have overlapping bands

Copper shows band overlap as a conductor

Use Doppler effect formula: Δλ/λ=v/c

Calculate v=c(Δλ/λ)=3×10⁸(1/600)

Convert to km/s: v=500 km/s