MHT CET 2025 19th April Evening Shift | Wave Optics Question 39 | Physics

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If the distance between the first minimum on either side of the central maximum is 1 cm , the wavelength of light used is

$2000 \mathop {\rm{A}}\limits^{\rm{^\circ }}$

$4000 \mathop {\rm{A}}\limits^ \circ$

$5000 \mathop {\rm{A}}\limits^ \circ$

$10000 \mathop {\rm{A}}\limits^ \circ$

MHT CET 2025 19th April Evening Shift

MCQ (Single Correct Answer)

-0

In Young's double slit experiment let 'd' be the distance between two slits and 'D' be the distance between the slits and the screen. Using a monochromatic source of wavelength ' $\lambda$ ', in an interference pattern, third minimum is observed exactly in front of one of the slits. If at the same point on the screen first minimum is to be obtained, the required change in the wavelength is [ $\mathrm{d} \& \mathrm{D}$ are not changed].

$2 \lambda$

$3 \lambda$

$4 \lambda$

$5 \lambda$

MHT CET 2025 19th April Morning Shift

MCQ (Single Correct Answer)

-0

In Young's double slit interference experiment, using two coherent sources of different amplitudes, the intensity ratio between bright to dark fringes is $5: 1$. The value of the ratio of resultant amplitudes of bright fringe to dark fringe is

$\left(\frac{\sqrt{5}+1}{\sqrt{5}-1}\right)$

$\sqrt{5}: 1$

$\left(\frac{\sqrt{5}-1}{\sqrt{5}+1}\right)$

$1: \sqrt{5}$

MHT CET 2025 19th April Morning Shift

MCQ (Single Correct Answer)

-0

In a Fraunhoffer diffraction, light of wavelength ' $\lambda$ ' is incident on slit of width ' d '. The diffraction pattern is observed on a screen placed at a distance ' $D$ '. The linear width of central maximum is equal to two times the width of the slit, then 'D' has value

$\frac{\mathrm{d}^2}{\lambda}$

$\frac{\mathrm{d}^2}{2 \lambda}$

$\frac{\mathrm{d}^2}{3 \lambda}$

$\frac{\mathrm{d}^2}{4 \lambda}$

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