Wave Optics — NEET Physics PYQ MCQs with Solutions

Solution: For the central maximum, the path difference is zero for both coherent sources in interference. Phase difference $= \frac{2\pi}{\lambda} \times$ path difference $= \frac{2\pi}{\lambda} \times 0 = 0$

Solution: When white light is incident on a soap bubble, it is partly reflected from the upper surface and partly from the lower surface. These two reflected beams superpose and cause interference. The colours that satisfy the condition of maxima are visible in the reflected light.

Solution: For a point in front of one slit, $y = \frac{d}{2}$. Path difference $= \frac{dy}{D} = \frac{d \times \frac{d}{2}}{10d} = \frac{d}{20} = \frac{\lambda}{4}$ (since $d = 5\lambda$) Phase difference $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} = 90°$ $I = I_0\cos^2\frac{\phi}{2} = I_0\cos^2 45° = I_0 \times \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2}$

Solution: With $\frac{I_2}{I_1} = n$: $\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{(\sqrt{I_2} + \sqrt{I_1})^2 - (\sqrt{I_2} - \sqrt{I_1})^2}{(\sqrt{I_2} + \sqrt{I_1})^2 + (\sqrt{I_2} - \sqrt{I_1})^2}$ $= \frac{(\sqrt{n} + 1)^2 - (\sqrt{n} - 1)^2}{(\sqrt{n} + 1)^2 + (\sqrt{n} - 1)^2} = \frac{4\sqrt{n}}{2(n + 1)} = \frac{2\sqrt{n}}{n + 1}$

Solution: Width of central maximum of the single-slit pattern = width of 10 double-slit maxima: $\frac{2D\lambda}{a} = 10\left(\frac{\lambda D}{d}\right)$ $a = \frac{2d}{10} = \frac{d}{5} = \frac{10^{-3}}{5} = 0.2 \times 10^{-3}$ m $a = 0.2$ mm

Solution: Fringe width $\beta = \frac{\lambda D}{d}$ With $D_2 = 2D_1$ and $d_2 = \frac{d_1}{2}$ (same $\lambda$): $\frac{\beta_2}{\beta_1} = \frac{D_2}{D_1} \times \frac{d_1}{d_2} = 2 \times 2 = 4$ The fringe width becomes four times.

Solution: In a YDSE, the path difference for the nth minimum is $\Delta y = (2n - 1)\frac{\lambda}{2}$ For the 5th minimum, $n = 5$: $\Delta y = [2(5) - 1]\frac{\lambda}{2} = \frac{9\lambda}{2}$

Solution: If the YDSE apparatus is immersed in a liquid of refractive index $\mu$, the wavelength of light — and hence the angular width — decreases $\mu$ times: $\theta' = \frac{\theta}{\mu} = \frac{0.2°}{4/3} = 0.15°$

Solution: Angular width of a fringe: $\theta = \frac{\lambda}{d}$, so $\theta \propto \frac{1}{d}$ $\frac{\theta_1}{\theta_2} = \frac{d_2}{d_1}$ $\frac{0.20°}{0.21°} = \frac{d_2}{2\text{ mm}}$ $d_2 = 2 \times \frac{0.20}{0.21} = \frac{0.40}{0.21} = 1.90$ mm

Solution: Position of 5th dark fringe in air = position of 8th bright fringe in the medium: $(2 \times 5 - 1)\frac{\lambda D}{2d} = 8\frac{\lambda D}{\mu d}$ $\frac{9}{2} = \frac{8}{\mu}$ $\mu = \frac{16}{9} = 1.777 \approx 1.78$

Solution: Intensity is proportional to slit width: $\frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{1}{25}$ $\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{I_2} + \sqrt{I_1}}{\sqrt{I_2} - \sqrt{I_1}}\right)^2 = \left(\frac{5 + 1}{5 - 1}\right)^2 = \left(\frac{6}{4}\right)^2 = \frac{36}{16} = \frac{9}{4}$

Solution: Net intensity $I = 4I_0\cos^2\frac{\phi}{2}$ For path difference $\lambda$: $\phi = 2\pi$, so $K = 4I_0$ For path difference $\frac{\lambda}{4}$: $\phi = \frac{\pi}{2}$, so $K' = 4I_0\cos^2\frac{\pi}{4} = 2I_0$ Comparing: $K' = \frac{K}{2}$

Solution: Fringe width for the first wavelength: $\beta_1 = \frac{\lambda_1 D}{d} = \frac{12000 \times 10^{-10} \times 2}{2 \times 10^{-3}} = 1.2$ mm Fringe width for the second wavelength: $\beta_2 = \frac{\lambda_2 D}{d} = \frac{10000 \times 10^{-10} \times 2}{2 \times 10^{-3}} = 1$ mm At 6 mm from the central bright fringe, the 5th fringe of the first wavelength coincides with the 6th fringe of the second wavelength.

Solution: Limit of resolution of a telescope: $d\theta = \frac{1.22\lambda}{d} = \frac{1.22 \times 600 \times 10^{-9}}{2} = 3.66 \times 10^{-7}$ rad

Solution: The Brewster's angle for an interface satisfies $\tan i_b = \mu$. Since $\mu > 1$ for any denser medium, $i_b$ lies between 45° and 90°.

Solution: Angular width of central maximum: $2\theta = \frac{2\lambda}{a}$, so $\theta \propto \lambda$ (slit width constant). New angular width $= 0.7\theta_0$, so $\lambda_2 = 0.7 \times \lambda_1 = 0.7 \times 6000 = 4200$ Å

Solution: Angular magnification $M = \frac{f_0}{f_e}$ — for large magnification the objective should have a large focal length $f_0$. Angular resolution $R = \frac{a}{1.22\lambda}$ — for high resolution the objective should have a large diameter a. Hence the objective lens should have a large focal length and a large diameter.

Solution: When the reflected and refracted rays are perpendicular to each other, the angle of incidence is the Brewster's angle ($\tan i = \mu$, not $\tan i = 1/\mu$). At this angle the reflected light is completely plane polarised, with its electric vector perpendicular to the plane of incidence.

Solution: Resolving power of a microscope $\propto \frac{1}{\lambda}$ $\frac{RP_1}{RP_2} = \frac{\lambda_2}{\lambda_1} = \frac{6000}{4000} = \frac{3}{2}$

Solution: After $P_1$: $I_1 = \frac{I_0}{2}$ After $P_3$ (at 45° to $P_1$): $I_2 = \frac{I_0}{2}\cos^2 45° = \frac{I_0}{4}$ After $P_2$ (at 45° to $P_3$): $I_3 = \frac{I_0}{4}\cos^2 45° = \frac{I_0}{8}$

Solution: First minimum: $a\sin 30° = \lambda \Rightarrow a = 2\lambda$ First secondary maximum: $a\sin\theta_1 = \frac{3\lambda}{2}$ $\sin\theta_1 = \frac{3\lambda}{2a} = \frac{3\lambda}{4\lambda} = \frac{3}{4}$ $\theta_1 = \sin^{-1}\left(\frac{3}{4}\right)$

Solution: The first minimum is formed at a distance $Y = \frac{\lambda D}{a}$, where D is the focal length of the lens. $Y = \frac{(5 \times 10^{-5}) \times 60}{0.02} = 0.15$ cm

Solution: For the first minimum, $\sin\theta = \frac{\lambda}{a}$, and for small angles $\sin\theta = \frac{Y}{D}$ $Y = \frac{\lambda D}{a}$ Width of central maximum $= 2Y = \frac{2\lambda D}{a}$

Solution: For the first minimum, $a\sin\theta = \lambda$. The path difference between the wavelet from the edge and from the midpoint of the slit is $\frac{a}{2}\sin\theta = \frac{\lambda}{2}$ Phase difference $= \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi$ rad

Solution: The distance between the first dark fringes on either side is the width of the central maximum: $x = \frac{2\lambda D}{a} = \frac{2 \times 600 \times 10^{-9} \times 2}{10^{-3}} = 2.4 \times 10^{-3}$ m $= 2.4$ mm

Solution: Angular resolution of a telescope: $= \frac{1.22\lambda}{d} = \frac{1.22 \times 5000 \times 10^{-10}}{10 \times 10^{-2}} = 6.1 \times 10^{-6} \approx 10^{-6}$ rad