Wave Optics

Answer: (c)

1. Newton's corpuscular theory predicts that light travels faster in a denser medium: $v_{glass} > v_{air}$.
2. It does explain straight-line travel, unchanged speed after reflection and a change of speed on refraction.
3. The statement $v_{air} > v_{glass}$ contradicts the corpuscular theory.
4. So the answer is (c).

Answer: (d)

1. Huygens' principle: every point on a wavefront acts as a source of secondary wavelets.
2. The forward envelope (common tangent) of these wavelets gives the new wavefront.
3. So the answer is (d).

Answer: (c)

1. On reflection the wave stays in the same medium, so speed is unchanged.
2. Hence angle of reflection = angle of incidence, $r = i$.
3. Widths: $AB = L\cos i$ and $CD = L\cos r$ for the same length $L$ of the surface.
4. So the answer is (c) $AB = CD$.

Answer: (c)

1. $\lambda_{medium} = \lambda_0/\mu$, so $\lambda_1\mu_1 = \lambda_2\mu_2$.
2. $540 \times \frac{4}{3} = \lambda_2 \times \frac{3}{2}$
3. $\lambda_2 = \dfrac{720}{1.5} = 480$ nm
4. So the answer is (c) 480 nm.

Answer: (c)

1. Statement I: a prism only deviates a plane wave, which stays plane; a pinhole acts like a point source, giving a spherical wave. True.
2. Statement II: a wider slit gives less diffraction, so the emerging wavefront is flatter and its curvature decreases. False.
3. So the answer is (c).

Answer: (a)

1. In the same time, light travels distance $z$ in air and distance $w$ in water.
2. So $\dfrac{z}{w} = \dfrac{c}{v_w} = {}_a\mu_w$ (water w.r.t. air).
3. Air w.r.t. water: ${}_w\mu_a = \dfrac{1}{{}_a\mu_w} = \dfrac{w}{z}$.
4. So the answer is (a) $\dfrac{w}{z}$.

Answer: (b)

1. In wave theory, colour depends on wavelength, so (a) is correct.
2. Wave theory predicts $v_{denser} < v_{rarer}$ and explains reflection and refraction, so (c) and (d) are correct.
3. Colour due to size of corpuscles belongs to Newton's corpuscular theory, not Huygens' theory.
4. So the answer is (b).

Answer: (b)

1. Wavelength in air: $\lambda_0 = c/\nu = \dfrac{3\times10^{8}}{5\times10^{14}} = 6\times10^{-7}$ m $= 600$ nm.
2. In the medium: $\lambda = \lambda_0/\mu = 600/2 = 300$ nm.
3. So the answer is (b) 300 nm.

Answer: (b)

1. For the plane $x + y + z =$ constant, the normal (direction of propagation) is $\hat n = \dfrac{\hat i + \hat j + \hat k}{\sqrt3}$.
2. Angle with the $x$-axis: $\cos\alpha = \hat n \cdot \hat i = \dfrac{1}{\sqrt3}$.
3. So the answer is (b) $\cos^{-1}(1/\sqrt3)$.

Answer: (b)

1. Doppler shift: $\dfrac{\Delta\lambda}{\lambda} = \dfrac{v}{c}$.
2. $\dfrac{\Delta\lambda}{\lambda} = 0.02\% = 2\times10^{-4}$
3. $v = 2\times10^{-4} \times 3\times10^{8} = 6\times10^{4}$ m/s $= 60$ km/s
4. So the answer is (b) 60 km/s.

Answer: (a)

1. A wavefront is the locus of points vibrating in the same phase.
2. Rays (direction of energy flow) are always normal to the wavefront.
3. So the answer is (a) perpendicular to the direction of propagation.

Answer: (c)

1. A point source at the focus gives a diverging spherical wavefront at the lens.
2. The convex lens turns these rays into a parallel beam.
3. Parallel rays correspond to a plane wavefront.
4. So the answer is (c) plane.

Answer: (c)

1. For the same length $L$ of the interface: incident width $= L\cos i$, refracted width $= L\cos r$.
2. Going from denser to rarer, the ray bends away from the normal: $r > i$.
3. So $\cos r < \cos i$ and the refracted wavefront is narrower.
4. So the answer is (c) decreases.

Answer: (d)

1. Refractive index of glass depends on wavelength ($\mu_v > \mu_r$), so focal length differs for each colour.
2. Violet focuses closer to the lens than red, so the colours do not meet at one point.
3. This defect is chromatic aberration, so the answer is (d).

Answer: (a)

1. Green just grazes the surface, so the angle of incidence equals the critical angle for green.
2. $\sin C = 1/\mu$; colours with smaller $\mu$ (longer $\lambda$) have a larger critical angle, so for them $i < C$ and they refract out.
3. Blue and violet have larger $\mu$, so $i > C$ and they are totally internally reflected.
4. So the answer is (a) yellow, orange and red.

Answer: (c)

1. Doppler shift: $v = c\dfrac{\Delta\lambda}{\lambda}$, with $\Delta\lambda = 670.7 - 670 = 0.7$ nm.
2. $v = 3\times10^{8} \times \dfrac{0.7}{670} \approx 3.13\times10^{5}$ m/s (galaxy receding).
3. So the answer is (c) $3.13\times10^{5}$ m/s.

Answer: 6

1. Doppler shift: $\Delta\lambda = \lambda\dfrac{v}{c}$.
2. $\Delta\lambda = 630\times10^{-9} \times \dfrac{286\times10^{3}}{3\times10^{8}} = 630\times10^{-9} \times 9.53\times10^{-4}$
3. $\Delta\lambda \approx 6.0\times10^{-10}$ m
4. So $x = 6$.

Answer: (a)

1. Doppler shift: $v = c\dfrac{\Delta\lambda}{\lambda}$, with $\Delta\lambda = 5896 - 5890 = 6$ Å.
2. $v = 3\times10^{8} \times \dfrac{6}{5890} \approx 3.06\times10^{5}$ m/s
3. $v \approx 306$ km/s
4. So the answer is (a) 306 km/s.

Answer: (d)

1. Frequency depends only on the source, so it does not change on entering glass.
2. Photon energy $E = h\nu$ depends only on frequency, so $E$ stays the same.
3. Speed, wavelength and momentum $p = h/\lambda$ all change.
4. So the answer is (d) energy.

Answer: (c)

1. For a source approaching the observer, $\Delta\lambda = -\lambda v/c$: wavelength decreases (blue shift).
2. As the star accelerates, $v$ keeps increasing, so the shift keeps growing.
3. Yellow light therefore shifts gradually towards blue.
4. So the answer is (c).

Answer: (b)

1. Refractive index increases with height, so the speed $v = c/\mu$ is smaller higher up.
2. The upper part of each wavefront moves slower than the lower part, so the wavefront tilts.
3. The beam (normal to the wavefront) turns towards the region of higher $\mu$.
4. So the answer is (b) bends upwards.

Answer: (a)

1. At sunrise and sunset sunlight travels a long path through the atmosphere.
2. Rayleigh scattering $\propto 1/\lambda^4$ removes blue light from the direct beam most strongly.
3. Mostly red light reaches the observer, so the answer is (a) scattering of light.

Answer: (d)

1. The beam is initially parallel, i.e. all rays are parallel to the axis.
2. Rays are normal to the wavefront, so parallel rays mean a plane wavefront.
3. The intensity-dependent $\mu$ changes the shape only later as the beam travels.
4. So the answer is (d) planar.

Answer: (b)

1. Speed in the material: $v = \nu\lambda = (2\times10^{14})(5000\times10^{-10}) = 1\times10^{8}$ m/s.
2. Refractive index $n = c/v = \dfrac{3\times10^{8}}{1\times10^{8}} = 3$.
3. So the answer is (b) 3.00.

Answer: (b)

1. Frequency is fixed by the source and does not change on refraction.
2. In water $v' = v/\mu$ and $\lambda' = \lambda/\mu$, and part of the light is reflected so $I' < I$.
3. So the answer is (b) $n = n'$.

Answer: (a)

1. Doppler shift for light ($v \ll c$): $\Delta\lambda = \lambda\dfrac{v}{c}$.
2. $\Delta\lambda = 5000 \times \dfrac{1.5\times10^{6}}{3\times10^{8}} = 5000 \times 5\times10^{-3}$
3. $\Delta\lambda = 25$ Å (a decrease, since the star approaches).
4. So the answer is (a) 25 Å.

Answer: (d)

1. Diffraction spreads light by an angle $\theta \approx \lambda/a$.
2. When the size $a \gg \lambda$, this spreading is negligible and light travels in straight lines (rays).
3. So the answer is (d) much larger than the wavelength.

Answer: (d)

1. Diffraction spreads light by an angle $\theta \approx \lambda/a$.
2. When the size $a \gg \lambda$, this spreading is negligible and light travels in straight lines (rays).
3. So the answer is (d) much larger than the wavelength.

Answer: (d)

1. Huygens' construction explains propagation effects: reflection, refraction, diffraction.
2. The origin of spectra comes from emission and absorption by atoms (quantum transitions), which wave construction cannot explain.
3. So the answer is (d) origin of spectra.

Answer: (c)

1. $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$ and $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$.
2. Best contrast needs perfectly dark fringes, $I_{min} = 0$, which happens only when $I_1 = I_2$.
3. So the answer is (c) equal.

Answer: (c)

1. A: energy taken from dark regions appears in bright regions, so total energy is conserved. True.
2. B: interference and diffraction are shown by all waves (sound, water waves, matter waves), not only light. False.
3. So the answer is (c).

Answer: (c)

1. Dark fringe: phase difference is an odd multiple of $\pi$: $\pi, 3\pi, 5\pi, \ldots$
2. For $n = 1, 2, 3, \ldots$ this is $\phi = (2n-1)\pi$.
3. So the answer is (c) $(2n-1)\pi$.

Answer: (c)

1. $\frac{I_{max}}{I_{min}} = \left(\frac{a_1+a_2}{a_1-a_2}\right)^2 = 36$
2. $\frac{a_1+a_2}{a_1-a_2} = 6 \Rightarrow a_1 + a_2 = 6a_1 - 6a_2 \Rightarrow 5a_1 = 7a_2$
3. $\frac{a_1}{a_2} = \frac{7}{5}$
4. So the answer is (c) $7 : 5$.

Answer: (d)

1. $\frac{I_{max}}{I_{min}} = \left(\frac{a_1+a_2}{a_1-a_2}\right)^2$
2. $= \left(\frac{3+1}{3-1}\right)^2 = 2^2 = 4$
3. So the answer is (d) $4 : 1$.

Answer: (c)

1. $I_{max} + I_{min} = 2(I_1 + I_2)$, $I_{max} - I_{min} = 4\sqrt{I_1 I_2}$
2. Ratio $= \frac{I_1 + I_2}{2\sqrt{I_1 I_2}}$; divide by $I_1$ with $\beta = \frac{I_2}{I_1}$:
3. $= \frac{1+\beta}{2\sqrt{\beta}}$
4. So the answer is (c) $\frac{\beta + 1}{2\sqrt{\beta}}$.

Answer: (b)

1. For equal slits $I = I_{max}\cos^2\dfrac{\phi}{2}$ with $\phi = \dfrac{2\pi}{\lambda}\Delta x$.
2. At $P$: $\Delta x = 0$, $\phi = 0$, $I_P = I_{max}$.
3. At $Q$: $\Delta x = \lambda/4$, $\phi = \pi/2$, $I_Q = I_{max}\cos^2 45^\circ = \dfrac{I_{max}}{2}$.
4. So the answer is (b) $2 : 1$.

Answer: (c)

1. $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$, with $2\sqrt{I \cdot 9I} = 6I$.
2. At $P$ ($\phi = \pi/2$): $I_P = I + 9I + 0 = 10I$.
3. At $Q$ ($\phi = \pi$): $I_Q = 10I - 6I = 4I$.
4. $I_P - I_Q = 6I$. So the answer is (c) $6I$.

Answer: (c)

1. Moving one fringe width $\beta$ changes path difference by $\lambda$, so $\beta/4$ gives $\Delta x = \lambda/4$, $\phi = \pi/2$.
2. $I = I_{max}\cos^2\dfrac{\phi}{2} = I_{max}\cos^2 45^\circ = \dfrac{I_{max}}{2}$
3. $\dfrac{I_{max}}{I} = 2$. So the answer is (c) $2$.

Answer: (d)

1. $\phi = \dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{6} = \dfrac{\pi}{3}$
2. $\dfrac{I}{I_0} = \cos^2\dfrac{\phi}{2} = \cos^2 30^\circ = \dfrac{3}{4}$
3. So the answer is (d) $0.75$.

Answer: (c)

1. Path difference $\lambda$ gives $\phi = 2\pi$, a bright fringe, so $X = I_{max}$.
2. For $\lambda/6$: $\phi = \pi/3$, $I = X\cos^2 30^\circ = \dfrac{3X}{4}$.
3. So the answer is (c) $\dfrac{3X}{4}$.

Answer: (a)

1. Amplitudes $a$ and $2a$, so intensities $I_0$ and $4I_0$; $I_m = (a + 2a)^2 \propto 9I_0$.
2. $I = I_0 + 4I_0 + 2\sqrt{4}I_0\cos\phi = 5I_0 + 4I_0\cos\phi$
3. Using $\cos\phi = 2\cos^2\dfrac{\phi}{2} - 1$: $I = I_0\left(1 + 8\cos^2\dfrac{\phi}{2}\right)$.
4. With $I_0 = I_m/9$: $I = \dfrac{I_m}{9}\left(1 + 8\cos^2\dfrac{\phi}{2}\right)$. So the answer is (a).

Answer: (b)

1. $I = I_{max}\cos^2\dfrac{\phi}{2}$, $\phi = \dfrac{2\pi}{\lambda}\Delta x$.
2. $\Delta x = 0$: $I_1 = I_{max}$. $\Delta x = \lambda/4$: $\phi = \pi/2$, $I_2 = I_{max}/2$.
3. So the answer is (b) $2 : 1$.

Answer: (b)

1. For equal slits $I = 4I_0\cos^2\dfrac{\phi}{2}$.
2. $\Delta x = \lambda/4$: $\phi = \pi/2$, $I_1 = 4I_0\cos^2 45^\circ = 2I_0$.
3. $\Delta x = \lambda/3$: $\phi = 2\pi/3$, $I_2 = 4I_0\cos^2 60^\circ = I_0$.
4. $\dfrac{I_1 + I_2}{I_0} = 3$. So the answer is (b) $3$.

Answer: (a)

1. Path difference $\lambda$ is a bright fringe, so $K = I_{max}$.
2. For $\lambda/3$: $\phi = 2\pi/3$, $I = K\cos^2 60^\circ = \dfrac{K}{4}$.
3. So the answer is (a) $\dfrac{K}{4}$.

Answer: (c)

1. $I = I_{max}\cos^2\dfrac{\phi}{2}$, $\phi = \dfrac{2\pi}{\lambda}\Delta x$.
2. At $A$: $\phi = 2\pi/3$, $I_A = I_{max}\cos^2 60^\circ = \dfrac{I_{max}}{4}$.
3. At $B$: $\phi = \pi/3$, $I_B = I_{max}\cos^2 30^\circ = \dfrac{3I_{max}}{4}$.
4. $\dfrac{I_A}{I_B} = \dfrac{1}{3}$. So the answer is (c) $1/3$.

Answer: (b)

1. In front of one slit, $y = \dfrac{d}{2} = 1\ \text{mm}$ from the centre.
2. Path difference $\Delta x = \dfrac{yd}{D} = \dfrac{(10^{-3})(2\times10^{-3})}{10} = 2\times10^{-7}\ \text{m} = \dfrac{\lambda}{3}$ (as $\lambda = 6\times10^{-7}$ m).
3. $\phi = \dfrac{2\pi}{3}$, so $I = 4I_o\cos^2 60^\circ = I_o$.
4. So the answer is (b) $I_o$.

Answer: 6

1. $I = I_{max}\cos^2\frac{\phi}{2} = \frac{3}{4}I_{max} \Rightarrow \cos\frac{\phi}{2} = \frac{\sqrt{3}}{2}$
2. $\frac{\phi}{2} = \frac{\pi}{6} \Rightarrow \phi = \frac{\pi}{3}$
3. Path difference $\Delta x = \frac{\lambda}{2\pi}\phi = \frac{\lambda}{6}$
4. So $x = 6$.

Answer: (c)

1. Each slit gives $I_0$; resultant $I = 4I_0\cos^2\frac{\phi}{2}$.
2. $4I_0\cos^2\frac{\phi}{2} = I_0 \Rightarrow \cos\frac{\phi}{2} = \frac{1}{2} \Rightarrow \phi = \frac{2\pi}{3}$
3. $\Delta x = \frac{\lambda}{2\pi}\cdot\frac{2\pi}{3} = \frac{\lambda}{3} = \frac{6000}{3} = 2000\ \text{Å}$
4. So the answer is (c) 2000.

Answer: (a)

1. At destructive interference the resultant amplitude is $|A_1 - A_2|$.
2. If $A_1 \ne A_2$, this is not zero, so dark fringes are not completely dark.
3. So the answer is (a).

Answer: (d)

1. Sustained interference needs coherent sources of the same frequency.
2. Red and green filters pass light of different frequencies, so the two slits are not coherent.
3. The phase difference changes rapidly and no steady fringes form; the screen is just uniformly illuminated.
4. So the answer is (d) there shall be no interference fringes.

Answer: (c)

1. Dark fringe: phase difference is an odd multiple of $\pi$: $\pi, 3\pi, 5\pi, \ldots$
2. For $n = 1, 2, 3, \ldots$ this is $\phi = (2n-1)\pi$.
3. So the answer is (c) $(2n-1)\pi$.

Answer: (b)

1. Intensity ratio $1:9 \Rightarrow$ amplitude ratio $1:3$.
2. $\frac{I_{max}}{I_{min}} = \left(\frac{1+3}{3-1}\right)^2 = \frac{16}{4} = 4$
3. So the answer is (b) $4 : 1$.

Answer: 24

1. $I_{max} = (\sqrt{4I}+\sqrt{9I})^2 = (2+3)^2 I = 25I$
2. $I_{min} = (\sqrt{9I}-\sqrt{4I})^2 = (3-2)^2 I = I$
3. $I_{max} - I_{min} = 24I$
4. So $x = 24$.

Answer: (b)

1. $I_{max} + I_{min} = 2(I_1 + I_2)$, $I_{max} - I_{min} = 4\sqrt{I_1 I_2}$
2. Ratio $= \frac{I_1 + I_2}{2\sqrt{I_1 I_2}}$; with $\frac{I_1}{I_2} = b$ this is $\frac{1+b}{2\sqrt{b}}$
3. So the answer is (b) $\frac{1+b}{2\sqrt{b}}$.