JEE Physics Wave Optics MCQs Set 08

Answer: (b) \( 7.692\pi \)

Explanation:
1. \( \Delta\phi=\frac{2\pi}{\lambda}\Delta x=\frac{2\pi\times2.1\times10^{-6}}{5.46\times10^{-7}} \).
2. \( =2\pi\times3.846=7.692\pi \) rad.

Teacher's Note:
1. Phase difference equals 2π times the path difference measured in wavelengths.
2. Convert units first: 2.1 micron = 2.1 × 10⁻⁶ m and 5460 Å = 5.46 × 10⁻⁷ m.

Answer: (b) \( \frac{\lambda}{2\pi}\left(\phi+\frac\pi2\right) \)

Explanation:
1. Write the second wave as a sine: \( \cos(\omega t+\phi)=\sin\left(\omega t+\phi+\frac\pi2\right) \).
2. Phase difference \( =\phi+\frac\pi2 \).
3. Path difference \( =\frac{\lambda}{2\pi}\times\text{phase difference}=\frac\lambda{2\pi}\left(\phi+\frac\pi2\right) \).

Teacher's Note:
1. Both waves must be expressed in the same trigonometric form before comparing phases.
2. Path difference = (λ/2π) × phase difference.

Answer: (c) 9I and I

Explanation:
1. Amplitudes are in the ratio \( \sqrt I:\sqrt{4I}=1:2 \).
2. \( I_{max}=(\sqrt I+2\sqrt I)^2=9I \) and \( I_{min}=(2\sqrt I-\sqrt I)^2=I \).

Teacher's Note:
1. For coherent beams the amplitudes add, not the intensities.
2. Imax = (√I1 + √I2)² and Imin = (√I1 − √I2)².

Answer: (c) its wavelength decreases but its frequency remains unchanged

Explanation:
1. Frequency is fixed by the source and does not change at a boundary.
2. Speed falls to \( v=\frac cn \), so \( \lambda'=\frac vf=\frac\lambda n \) decreases.

Teacher's Note:
1. The number of wave crests arriving per second must equal the number leaving, so frequency is unchanged.
2. In a denser medium the speed is lower, so the wavelength is shorter.

Answer: (b) 67

Explanation:
1. The fringe width is \( \beta\propto\lambda \), so the number of fringes in a fixed field is proportional to \( \frac1\lambda \).
2. \( N_2=62\times\frac{5893}{5461}=66.9\approx67 \).

Teacher's Note:
1. A smaller wavelength gives narrower fringes, so more of them fit into the same field of view.
2. N1β1 = N2β2 gives N2 = N1λ1/λ2.

Answer: (c) \( (2n-1)\frac{\lambda D}{2d} \)

Explanation:
1. Dark fringes occur where the path difference is an odd number of half wavelengths: \( \frac{yd}{D}=(2n-1)\frac\lambda2 \).
2. \( y_n=(2n-1)\frac{\lambda D}{2d} \), with n = 1, 2, 3, ...

Teacher's Note:
1. The first dark fringe is at half a fringe width from the centre, the second at one and a half fringe widths, and so on.
2. Bright fringes are at nλD/d, dark fringes at (2n − 1)λD/2d.

Answer: (a) \( 1.5\times10^{10}\ \text{Hz} \)

Explanation:
1. Incident and reflected waves form a standing wave; successive maxima are \( \frac\lambda2 \) apart.
2. 14 maxima have 13 gaps: \( 13\times\frac\lambda2=0.13\Rightarrow\lambda=0.02\ \text{m} \).
3. \( f=\frac c\lambda=\frac{3\times10^8}{0.02}=1.5\times10^{10}\ \text{Hz} \).

Teacher's Note:
1. Count intervals, not maxima: n maxima span (n − 1) intervals.
2. Adjacent antinodes of a standing wave are λ/2 apart.

Answer: (c) \( \frac{D\lambda}{3d} \)

Explanation:
1. If one slit alone gives intensity \( I_0 \), the double-slit intensity is \( I=4I_0\cos^2\frac\phi2 \).
2. Setting \( I=I_0\Rightarrow\cos^2\frac\phi2=\frac14\Rightarrow\frac\phi2=\frac\pi3\Rightarrow\phi=\frac{2\pi}{3} \).
3. Path difference \( =\frac\lambda3=\frac{yd}{D}\Rightarrow y=\frac{D\lambda}{3d} \).

Teacher's Note:
1. The central maximum has four times the single-slit intensity, so we need the point where cos²(φ/2) falls to 1/4.
2. Phase difference 2π/3 corresponds to a path difference of λ/3.

Answer: (d) 5600

Explanation:
1. Position of the 4th bright fringe of 6300 Å: \( y=4\times6300\,\frac Dd \).
2. Position of the 5th dark fringe of λ: \( y=\left(5-\tfrac12\right)\lambda\,\frac Dd=4.5\lambda\,\frac Dd \).
3. Equal positions: \( 4.5\lambda=25200\Rightarrow\lambda=5600\ \text{Å} \).

Teacher's Note:
1. The n-th bright fringe is at nβ and the n-th dark fringe at (n − ½)β.
2. Coincidence means the same distance from the centre for both wavelengths.

Answer: (a) 2

Explanation:
1. A point at a quarter of the fringe width from the centre has path difference \( \frac\lambda4 \), so phase difference \( \frac\pi2 \).
2. \( I=I_0\cos^2\frac\phi2=I_0\cos^2\frac\pi4=\frac{I_0}{2} \).
3. Ratio \( \frac{I_0}{I_0/2}=2 \).

Teacher's Note:
1. One fringe width corresponds to a path difference of λ (a phase of 2π).
2. The intensity varies as cos²(φ/2).

Answer: (b) 5

Explanation:
1. Deviation by the biprism: \( \delta=(\mu-1)\alpha=0.5\times2\times10^{-3}=10^{-3}\ \text{rad} \).
2. Separation of the virtual sources: \( d=2a\delta=2(1)(10^{-3})=2\times10^{-3}\ \text{m} \).
3. Fringe width: \( \beta=\frac{\lambda(a+b)}{d}=\frac{6\times10^{-7}\times5}{2\times10^{-3}}=1.5\times10^{-3}\ \text{m} \).
4. Width of the overlap region on the screen: \( 2b\delta=2(4)(10^{-3})=8\times10^{-3}\ \text{m} \).
5. Number of fringes \( =\frac{8}{1.5}=5.3\approx5 \).

Teacher's Note:
1. Fringes appear only in the region where the two refracted beams overlap, whose width is 2bδ.
2. Divide the overlap width by the fringe width.

Answer: (b) 12 cm

Explanation:
1. In the displacement method for a biprism: \( d=\sqrt{d_1d_2} \).
2. \( d=\sqrt{16\times9}=12\ \text{cm} \).

Teacher's Note:
1. The two lens positions give magnified and diminished images, d1 and d2, of the same source separation.
2. Geometric mean of the two image separations gives d.

Answer: (d) 180° change of phase in one of the reflected waves

Explanation:
1. On reflection from the denser medium a wave undergoes a phase change of \( \pi \) (180°), equal to an extra half wavelength of path.
2. So the condition for brightness in reflected light becomes the condition for darkness in transmitted light, and vice versa.

Teacher's Note:
1. Reflected and transmitted patterns are always complementary because energy is conserved.
2. The 180° phase jump on reflection at one surface is what interchanges bright and dark.

Answer: (b) black

Explanation:
1. For a very thin film the path difference is almost zero, but one reflection involves a phase change of π.
2. So the two reflected rays are out of phase for every wavelength and cancel: destructive interference.
3. Almost no light is reflected, so the film appears black.

Teacher's Note:
1. This is why the top of a vertical soap film turns black just before it breaks.
2. The 180° phase change occurs at the air-film surface only.

Answer: (a) 0.047 cm

Explanation:
1. First minimum of a single slit: \( a\sin\theta=\lambda \), and on the focal plane \( y=f\tan\theta\approx f\theta=\frac{\lambda f}{a} \).
2. \( y=\frac{5.896\times10^{-7}\times0.40}{5\times10^{-4}}=4.7\times10^{-4}\ \text{m}=0.047\ \text{cm} \).

Teacher's Note:
1. In single-slit diffraction the minima occur at a sin θ = nλ.
2. The lens brings the pattern to its focal plane, so use f as the screen distance.

Answer: (c) 37.5%

Explanation:
1. The first polaroid transmits half of unpolarised light: \( I_1=\frac{I_0}{2} \).
2. Malus' law for the second: \( I_2=I_1\cos^230°=\frac{I_0}{2}\times\frac34=\frac{3I_0}{8} \).
3. That is 37.5% of the incident light.

Teacher's Note:
1. Unpolarised light loses exactly half its intensity in the first polaroid.
2. Malus' law: I = I0 cos²θ.

Answer: (d) \( \tan^{-1}(n) \)

Explanation:
1. Brewster's law: reflected light is completely plane polarised when \( \tan i_B=n \).
2. So \( i_B=\tan^{-1}(n) \).

Teacher's Note:
1. At Brewster's angle the reflected and refracted rays are perpendicular to each other.
2. Tangent, not sine, appears in Brewster's law.

Answer: (b) the reflected ray will be completely polarised and the transmitted ray will be partially polarised

Explanation:
1. At the polarising (Brewster) angle, the reflected light has its electric vector perpendicular to the plane of incidence only, so it is completely polarised.
2. The refracted ray still contains the whole parallel component and some of the perpendicular component, so it is only partially polarised.

Teacher's Note:
1. Only a small fraction of the perpendicular component is reflected; most of it is transmitted.
2. So the transmitted beam is a mixture and is partially polarised.

Answer: (c) 90°

Explanation:
1. At Brewster's angle \( \tan\phi=\mu \), so \( \sin\phi=\mu\cos\phi \).
2. Snell's law: \( \sin\phi=\mu\sin r\Rightarrow\sin r=\cos\phi\Rightarrow r=90°-\phi \).
3. Angle between reflected and refracted rays \( =180°-\phi-r=90° \).

Teacher's Note:
1. The reflected and refracted rays are always at right angles at the polarising angle.
2. Use i + r = 90° at Brewster's angle.

Answer: (c) the top portion appears dark while the first colour to be observed as one moves down is violet

Explanation:
1. At the top the film is almost of zero thickness, and the phase change of π on one reflection gives destructive interference for every wavelength, so the top is dark.
2. Lower down, constructive interference in reflection needs \( 2\mu t=\left(n+\tfrac12\right)\lambda \).
3. The smallest thickness that satisfies this is for the smallest wavelength, so violet appears first as the thickness increases.

Teacher's Note:
1. For a wedge of increasing thickness, the colours appear in order of increasing wavelength: violet first, red last.
2. The order does not depend on the refractive index of the solution.