JEE Physics Wave Optics MCQs Set 10

Answer: (c) the fringe width will remain the same but the fringe pattern will shift down

Explanation:
1. The fringe width is \( \beta=\frac{\lambda D}{d} \), which does not depend on the source position.
2. With the source nearer the upper slit, the path S to S1 is smaller than S to S2. The central maximum needs equal total path, so the screen path from the lower slit must be shorter: the pattern shifts towards the lower slit, that is down.

Teacher's Note:
1. Moving the source does not change d, D or λ.
2. The central maximum always moves opposite to the source shift.

Answer: (b) B and D only

Explanation:
1. A stationary pattern needs the same frequency and a constant phase difference.
2. Equal amplitude or intensity only affects the contrast, not the existence of the pattern.

Teacher's Note:
1. Coherence means equal frequency and constant phase difference.
2. Unequal amplitudes just give fringes that do not reach zero.

Answer: (d) A and D only

Explanation:
1. The round trip path in the film is 2t = 2 wavelengths in the film, so the path alone gives constructive reflection.
2. If μ1 < μ2 < μ3 both reflections are from a denser medium, so there is no relative phase change: reflection is maximum (A true). If μ1 < μ2 > μ3 the relative change is π: reflection is minimum (B false).
3. Transmission is maximum when reflection is minimum. For μ1 > μ2 > μ3 there is no relative change, so reflection is maximum and transmission is minimum (C false). For μ1 > μ2 < μ3 the relative change is π, so transmission is maximum (D true).

Teacher's Note:
1. Count phase changes of π at each reflection from a denser medium.
2. Transmitted and reflected intensity are complementary.

Answer: (b) 81 : 1

Explanation:
1. Ray AB: \( I_1=0.2I \). Ray A'B': transmitted 0.8I, reflected 20% at the lower face gives 0.16I, transmitted again at the top gives \( I_2=0.8\times0.16I=0.128I \).
2. Amplitude ratio \( \frac{a_1}{a_2}=\sqrt{\frac{0.2}{0.128}}=1.25 \).
3. \( \frac{I_{max}}{I_{min}}=\left(\frac{a_1+a_2}{a_1-a_2}\right)^2=\left(\frac{2.25}{0.25}\right)^2=81 \).

Teacher's Note:
1. Intensity is proportional to the square of amplitude.
2. Track the fraction transmitted or reflected at each surface.

Answer: (a) A and C bright, B and D dark

Explanation:
1. On the perpendicular bisector the path difference is zero, so A and C are bright.
2. At B and D the path difference equals the separation, \( 5\ \mu m=2.5\lambda \), which is a half-integer number of wavelengths, so B and D are dark.

Teacher's Note:
1. Zero path difference gives a bright point.
2. A half-odd multiple of λ gives a dark point.

Answer: (c) 141

Explanation:
1. For reflected light, bright fringes occur where \( 2t=\left(m-\tfrac12\right)\lambda \). At the wire end \( 2t=0.096\ \text{mm}=96000\ \text{nm} \).
2. \( m-\tfrac12\le\frac{96000}{680}=141.2 \), so \( m\le141.7 \) and there are 141 bright fringes.

Teacher's Note:
1. The path difference in an air wedge is 2t plus a half-wave phase change.
2. Count the integers m allowed up to the wire.

Answer: (a) 9 : 1

Explanation:
1. \( \frac{I_{max}}{I_{min}}=\left(\frac{a_1+2a_1}{2a_1-a_1}\right)^2=\frac{9}{1} \).

Teacher's Note:
1. Maximum amplitude is the sum, minimum the difference.
2. Intensity is the square of amplitude.

Answer: (d) 0.225 mm

Explanation:
1. In water \( \lambda'=\frac{600}{4/3}=450\ \text{nm} \).
2. \( \beta=\frac{\lambda' D}{d}=\frac{450\times10^{-9}\times1}{2\times10^{-3}}=0.225\ \text{mm} \).

Teacher's Note:
1. Wavelength in a medium is λ divided by μ.
2. Use β = λD/d with the new wavelength.

Answer: (c) 18

Explanation:
1. Number of fringes in a fixed segment is inversely proportional to the fringe width, hence to λ.
2. \( N_2=12\times\frac{600}{400}=18 \).

Teacher's Note:
1. Fringe width is proportional to λ.
2. A fixed segment holds more fringes when each is narrower.

Answer: (a) 0.2 mm

Explanation:
1. \( I=I_{max}\cos^2\frac\phi2=0.75I_{max} \) gives \( \frac\phi2=30^\circ \), so \( \phi=\frac\pi3 \) and path difference \( \frac\lambda6 \).
2. \( y=\frac{\lambda D}{6d}=0.1\ \text{mm} \) on each side of the centre, so the two points nearest each other, on opposite sides, are \( 0.2\ \text{mm} \) apart.

Teacher's Note:
1. 75% intensity corresponds to cos²(φ/2) = 3/4.
2. The two nearest points lie symmetrically about the central maximum.

Answer: (c) 0.15 mm

Explanation:
1. The slit and its image are \( d=2\ \text{mm} \) apart and the reflection adds a phase change π, so the fringe pattern is shifted by half a fringe.
2. Fringe width \( \beta=\frac{\lambda D}{d}=\frac{600\times10^{-9}\times1}{2\times10^{-3}}=0.3\ \text{mm} \), so the first maximum is at \( \frac\beta2=0.15\ \text{mm} \).

Teacher's Note:
1. The mirror surface itself is a dark fringe.
2. The first maximum lies half a fringe width above the mirror.

Answer: (b) \( 1.98\times10^{-2}\ \text{mm} \)

Explanation:
1. The first minimum occurs where \( d\sin\theta=\frac\lambda2 \).
2. \( d=\frac{520\times10^{-9}}{2\sin0.75^\circ}=\frac{520\times10^{-9}}{2\times0.01309}=1.98\times10^{-5}\ \text{m} \).

Teacher's Note:
1. The first minimum has path difference λ/2.
2. 0.75° = 0.01309 rad.

Answer: (a) 5000 Å

Explanation:
1. The 9th bright fringe is at \( 9\beta \) and the 2nd dark fringe at \( 1.5\beta \).
2. So \( 7.5\beta=7.5\ \text{mm} \) gives \( \beta=1\ \text{mm} \).
3. \( \lambda=\frac{\beta d}{D}=\frac{10^{-3}\times0.5\times10^{-3}}{1}=5\times10^{-7}\ \text{m} \).

Teacher's Note:
1. Bright fringes lie at nβ and dark fringes at (n − ½)β.
2. Then λ = βd/D.

Answer: (c) \( 1.8\times10^{-2}\ \text{cm} \)

Explanation:
1. \( \beta=\frac{\lambda D}{d}=\frac{5.4\times10^{-5}\times80}{0.24}=1.8\times10^{-2}\ \text{cm} \).

Teacher's Note:
1. The two images act as coherent sources of separation 0.24 cm.
2. Use consistent units (cm).

Answer: (a) 1 mm

Explanation:
1. Image distance \( \frac1v=\frac1{0.1}-\frac1{0.15} \) gives \( v=0.30\ \text{m} \).
2. Each half has its optical centre 0.25 mm from the axis. The ray through that centre shows the image is at \( 0.25\times\frac{0.45}{0.15}=0.75\ \text{mm} \) from the axis, so \( d=1.5\ \text{mm} \).
3. Distance from the images to the screen \( D=1.30-0.30=1.0\ \text{m} \), so \( \beta=\frac{\lambda D}{d}=\frac{500\times10^{-9}\times1}{1.5\times10^{-3}}=0.33\ \text{mm} \) and the third maximum is at \( 3\beta=1\ \text{mm} \).

Teacher's Note:
1. Locate the two real images first; they are the coherent sources.
2. Measure D from the images, not from the lenses.

Answer: (d) increase

Explanation:
1. A smaller gap brings the optical centres closer to the axis, so the image separation d decreases.
2. Since \( \beta=\frac{\lambda D}{d} \), a smaller d gives a larger fringe width, so the distance OA increases.

Teacher's Note:
1. Image separation is proportional to the gap.
2. Fringe width is inversely proportional to d.

Answer: (c) \( \frac{\lambda}{2(\mu-1)\alpha} \)

Explanation:
1. Each half of the biprism deviates the beam by \( \delta=(\mu-1)\alpha \), so the two beams cross at an angle \( 2\delta \).
2. Two plane waves crossing at angle 2δ give fringe width \( \beta=\frac{\lambda}{2\delta}=\frac{\lambda}{2(\mu-1)\alpha} \), independent of D.

Teacher's Note:
1. For a parallel beam the sources are at infinity, so D does not appear.
2. Two plane waves at angle θ give β = λ/θ.

Answer: (a) \( \frac\pi6 \)

Explanation:
1. Write the second wave as a cosine: \( \sin\left(\theta+\frac\pi3\right)=\cos\left(\theta+\frac\pi3-\frac\pi2\right)=\cos\left(\theta-\frac\pi6\right) \).
2. Compared with \( \cos\theta \), the phase difference is \( \frac\pi6 \).

Teacher's Note:
1. Convert sine to cosine by subtracting π/2 from the argument.
2. Compare arguments of the two cosines.

Answer: (c) \( \frac{2\pi\mu x}{\lambda} \)

Explanation:
1. Wavelength in the medium is \( \lambda'=\frac\lambda\mu \).
2. Phase difference \( =\frac{2\pi x}{\lambda'}=\frac{2\pi\mu x}{\lambda} \).

Teacher's Note:
1. Equivalent optical path is μx.
2. Phase = 2π × (path)/λ.

Answer: (b) 6000 Å

Explanation:
1. \( \Delta\beta=\frac{\lambda\,\Delta D}{d} \), so \( \lambda=\frac{\Delta\beta\,d}{\Delta D}=\frac{3\times10^{-5}\times10^{-3}}{5\times10^{-2}}=6\times10^{-7}\ \text{m} \).

Teacher's Note:
1. Change in fringe width is λ times change in D divided by d.
2. 6 × 10⁻⁷ m = 6000 Å.

Answer: (a) zero and 1.5 mm

Explanation:
1. Extra optical path \( (\mu-1)t=0.5\times1.5\times10^{-6}=0.75\ \mu m=1.5\lambda \), a phase of 3π, which is a minimum, so the intensity is zero.
2. Shift \( =\frac{(\mu-1)tD}{d}=\frac{0.75\times10^{-6}\times1}{5\times10^{-4}}=1.5\ \text{mm} \).

Teacher's Note:
1. An odd number of half wavelengths of extra path gives darkness.
2. Shift = (μ − 1)tD/d.

Answer: (c) \( 1.575\ \mu m \)

Explanation:
1. The extra optical path is \( (1.53-1.33)t \) and for a minimum on the axis it must equal \( \frac{\lambda_{air}}{2} \).
2. \( t=\frac{630\times10^{-9}/2}{0.2}=1.575\times10^{-6}\ \text{m} \).

Teacher's Note:
1. The sheet is immersed in the liquid, so the relative index is 1.53 − 1.33 in optical path terms.
2. Minimum on axis needs λ/2 of extra path.