JEE Physics Wave Optics MCQs Set 03

Answer: (d) Polarisation

Explanation:
1. Huygens assumed longitudinal waves in ether, which cannot show polarisation.
2. Reflection, refraction and diffraction follow from the wavelet construction.

Teacher's Note:
1. Polarisation needs transverse waves.
2. Huygens' principle gives reflection and refraction.

Answer: (d) All of the above waves

Explanation:
1. Huygens' construction of secondary wavelets applies to any wave motion.
2. It is not limited to light or mechanical waves.

Teacher's Note:
1. It is a geometrical construction for wavefronts.

Answer: (d) All of the above

Explanation:
1. The wavelet construction gives the shape of the wavefront at later times.
2. Reflection and refraction laws follow from it.

Teacher's Note:
1. Each point on a wavefront is a source of secondary wavelets.

Answer: (c) The wavefront geometrically

Explanation:
1. The theory is a geometrical method to locate a new wavefront.
2. It does not directly give speed, wavelength or magnification.

Teacher's Note:
1. New wavefront is the envelope of secondary wavelets.

Answer: (c) Light exhibits interference

Explanation:
1. Interference can be explained only by waves.
2. Rectilinear motion, reflection and refraction are also explained by particles, and the photoelectric effect supports the particle nature.

Teacher's Note:
1. Interference, diffraction and polarisation prove the wave nature.

Answer: (a) Interference

Explanation:
1. Interference needs superposition of waves.
2. The photoelectric effect supports the particle nature.

Teacher's Note:
1. Interference and diffraction need waves.

Answer: (a) Is conserved, gets redistributed

Explanation:
1. Energy is neither created nor destroyed.
2. It moves from dark fringes to bright fringes and the average is unchanged.

Teacher's Note:
1. Conservation of energy holds in interference.

Answer: (c) 1 : 1

Explanation:
1. Visibility is best when \( I_{min}=0 \), which needs equal intensities.
2. Unequal intensities make dark fringes less dark.

Teacher's Note:
1. Equal intensity gives perfect contrast.

Answer: (d) \( 2\pi \)

Explanation:
1. Constructive interference needs \( \phi=2n\pi \).
2. Of the options only \( 2\pi \) qualifies.

Teacher's Note:
1. Even multiples of pi give bright fringes.

Answer: (c) By both division of amplitude and division of wavefront

Explanation:
1. Young's double slit uses division of wavefront and thin films use division of amplitude.
2. Both give coherent beams.

Teacher's Note:
1. Two ways: wavefront and amplitude.

Answer: (d) 4I

Explanation:
1. \( I_R=I+4I+2\sqrt{4}I\cos\phi \).
2. At A, \( 5I \); at B, \( 9I \).
3. The difference is 4I.

Teacher's Note:
1. Use I1 + I2 + 2 root(I1 I2) cos phi.

Answer: (b) 4I

Explanation:
1. \( I_{max}=(\sqrt I+\sqrt I)^2=4I \).
2. Amplitudes add, so intensity becomes four times.

Teacher's Note:
1. Constructive interference of equal sources gives 4I.

Answer: (c) 9 : 4

Explanation:
1. Amplitude ratio is 5 : 1.
2. \( I_{max}/I_{min}=(5+1)^2/(5-1)^2=36/16=9/4 \).

Teacher's Note:
1. Take the square root of the intensity ratio first.

Answer: (b) Both waves are incoherent

Explanation:
1. The sources are independent, so their phase relation is not fixed.
2. The waves are therefore incoherent even though the frequencies are equal.

Teacher's Note:
1. Independent sources are always incoherent.

Answer: (a) Have constant phase difference

Explanation:
1. A constant phase difference (coherence) gives a stationary pattern.
2. Equal amplitude only improves contrast.

Teacher's Note:
1. Coherence is the essential condition.

Answer: (d) 9I and I

Explanation:
1. \( I_{max}=(\sqrt I+2\sqrt I)^2=9I \).
2. \( I_{min}=(2\sqrt I-\sqrt I)^2=I \).

Teacher's Note:
1. Add and subtract amplitudes.

Answer: (d) \( \text{lags the second by }\pi/2 \)

Explanation:
1. \( \cos\omega t=\sin(\omega t+\pi/2) \), so the second wave is ahead by \( \pi/2 \).
2. The first wave lags by \( \pi/2 \).

Teacher's Note:
1. Cosine leads sine by pi/2.

Answer: (c) 9 : 4

Explanation:
1. \( (r+1)^2/(r-1)^2=25 \) gives \( (r+1)/(r-1)=5 \), so \( r=1.5 \) for amplitudes.
2. Intensity ratio \( =2.25=9:4 \).

Teacher's Note:
1. Solve for amplitude ratio then square.

Answer: (d) 7 : 5

Explanation:
1. \( (a_1+a_2)/(a_1-a_2)=6 \) gives \( a_1+a_2=6a_1-6a_2 \), so \( 5a_1=7a_2 \).
2. Hence \( a_1:a_2=7:5 \).

Teacher's Note:
1. Take the square root of the intensity ratio.
2. Then solve the linear equation.

Answer: (b) 0.1 μm

Explanation:
1. Thin film colours need thickness comparable to the wavelength of light (a fraction of a micrometre).
2. 10 nm is far too thin and 1 mm is too thick.

Teacher's Note:
1. Film thickness is of the order of lambda.

Answer: (b) Constant phase difference

Explanation:
1. A constant phase difference is the essential property.
2. Same colour (frequency) is needed too but alone is not enough.

Teacher's Note:
1. Coherence means constant phase relation.

Answer: (c) 4 : 1

Explanation:
1. Amplitude ratio is 3 : 1.
2. \( I_{max}/I_{min}=(3+1)^2/(3-1)^2=16/4=4 \).

Teacher's Note:
1. Same method as for 25 : 1.