# NEET and AIPMT NEET Physics Wave Optics MCQ Questions Solved

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NEET - 2017

Young's double slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. The refractive index of the medium is nearly

(a)1.25

(b) 1.59

(c) 1.69

(d) 1.78

(d) According to question, 5th fringe in air = 8th bright fringe in the medium

$\left(2×5-1\right)\frac{\lambda D}{2d}=8\frac{\lambda D}{\mu d}$

$9\frac{\lambda D}{2d}=8\frac{\lambda D}{\mu d}$

$\therefore$ Refractive index of the medium.

$\mu =\frac{16}{9}=1.7777=1.78$

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NEET - 2017

Two polarids  are placed with their axis perpendicular to each other. Unpolarised light ${I}_{o}$ is incident on ${P}_{1}$. A third polaroid ${P}_{3}$ is kept in between  such that its axis makes an angle ${45}^{°}$ with that of ${P}_{1}$. The intensity of transmitted light through ${P}_{2}$

(a)$\frac{{I}_{o}}{2}$

(b) $\frac{{I}_{o}}{4}$

(c) $\frac{{I}_{o}}{8}$

(d) $\frac{{I}_{o}}{16}$

(c) According to the question

From the above diagram, Intensity transmitted through${P}_{3}$

${I}_{2}=\frac{{I}_{o}}{2}{\mathrm{cos}}^{2}{45}^{°}$

Similarly, intensity transmitted through ${P}_{2}$

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NEET - 2016

The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio $\frac{{I}_{max}-{I}_{min}}{{I}_{max}+{I}_{min}}$ will be

(a) $\frac{\sqrt{n}}{n+1}$             (b) $\frac{2\sqrt{n}}{n+1}$

(c) $\frac{\sqrt{n}}{{\left(n+1\right)}^{2}}$         (d) $\frac{2\sqrt{n}}{{\left(n+1\right)}^{2}}$

(b) It is given that $\frac{{l}_{2}}{{l}_{1}}=n⇒{l}_{2}=n{l}_{1}$

Ratio of intensites is given by

$\frac{{l}_{max}-{l}_{min}}{{l}_{max}+{l}_{min}}=\frac{{\left(\sqrt{{l}_{2}}+\sqrt{{l}_{1}}\right)}^{2}_{\left(\sqrt{{l}_{2}}-\sqrt{{l}_{1}}\right)}^{2}}{{\left(\sqrt{{l}_{1}}+\sqrt{{l}_{2}}\right)}^{2}+{\left(\sqrt{{l}_{2}}-\sqrt{{l}_{1}}\right)}^{2}}$

$=\frac{{\left(\sqrt{\frac{{l}_{2}}{{l}_{1}}}+1\right)}^{2}-{\left(\sqrt{\frac{{l}_{2}}{{l}_{1}}-1}\right)}^{2}}{{\left(\sqrt{\frac{{l}_{2}}{{l}_{1}}}+1\right)}^{2}+{\left(\sqrt{\frac{{l}_{2}}{{l}_{1}}-1}\right)}^{2}}$

$=\frac{{\left(\sqrt{n}+1\right)}^{2}-{\left(\sqrt{n}-1\right)}^{2}}{{\left(\sqrt{n}+1\right)}^{2}+{\left(\sqrt{n}-1\right)}^{2}}=\frac{2\sqrt{n}}{n+1}$

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