MHT CET 2025 26th April Evening Shift | Wave Optics Question 4 | Physics

Graph shows the variation of fringe width ( X ) versus distance of the screen from the plane of the slits (D) in Young's double slit experiment. (keeping other parameters same, $d=$ distance between the slits). The wavelength of light used can be calculated as

MHT CET 2025 26th April Evening Shift Physics - Wave Optics Question 1 English

slope $\times \mathrm{d}^2$

$\frac{\mathrm{d}}{\text { slope }}$

$\frac{\text { slope }}{\mathrm{d}}$

slope $\times \mathrm{d}$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

In Young's double slit experiment, in an interference pattern second minimum is observed exactly in front of one slit. The distance between the two coherent sources is ' $d$ ' and the distance between source and screen is ' $D$ '. The wavelength of light source used is

$\frac{d^2}{4 D}$

$\frac{d^2}{3 D}$

$\frac{d^2}{2 D}$

$\frac{d^2}{D}$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The polarising angle of transparent medium is ' $\theta$ '. Let the speed of light in the medium be ' v '. Then the relation between ' $\theta$ ' and ' $\mathbf{v}$ ' is [ $\mathrm{c}=$ velocity of light in air]

$\quad \theta=\sin ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$

$\theta=\tan ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$

$\theta=\cot ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$

$\quad \theta=\cos ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$

MHT CET 2025 26th April Morning Shift

MCQ (Single Correct Answer)

-0

The ratio of the distance of $n^{\text {th }}$ bright band and $\mathrm{m}^{\text {th }}$ dark band from the central bright band in an interference pattern is

$n: m$

$m: n$

$n:\left(m-\frac{1}{2}\right)$

$\left(n-\frac{1}{2}\right): m$

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