Huygen's Principle and Interference of Light — 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: 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: 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: 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: 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.