1
MHT CET 2025 19th April Morning Shift
MCQ (Single Correct Answer)
+1
-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
A
$\frac{\mathrm{d}^2}{\lambda}$
B
$\frac{\mathrm{d}^2}{2 \lambda}$
C
$\frac{\mathrm{d}^2}{3 \lambda}$
D
$\frac{\mathrm{d}^2}{4 \lambda}$
2
MHT CET 2025 19th April Morning Shift
MCQ (Single Correct Answer)
+1
-0

Three identical polaroids $P_1, P_2$ and $P_3$ are placed one after another. The pass axis of $P_2$ and $\mathrm{P}_3$ are inclined at an angle of $60^{\circ}$ and $90^{\circ}$ with respect to axis of $\mathrm{P}_1$. The source has an intensity $256 \mathrm{~W} / \mathrm{m}^2$. The intensity of light at point ' O ' is $\left(\cos 30^{\circ}=\sqrt{3} / 2, \cos 60^{\circ}=0.5\right)$

MHT CET 2025 19th April Morning Shift Physics - Wave Optics Question 21 English

A
$24 \mathrm{~W} / \mathrm{m}^2$
B
$20 \mathrm{~W} / \mathrm{m}^2$
C
$16 \mathrm{~W} / \mathrm{m}^2$
D
$8 \mathrm{~W} / \mathrm{m}^2$
3
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In a single slit diffraction experiment, for a wavelength of light ' $\lambda$ ', half-angular width of the principle maxima is ' $\theta$ '. Also for wavelength of light $\mathrm{p} \lambda$, the half angular width of the principle maxima is $q \theta$. The ratio of the halfangular widths of the first secondary maxima in the first case to second case will be

A
$\mathrm{p}: 1$
B
$\mathrm{q}: 1$
C
$\mathrm{p}: \mathrm{q}$
D
$\mathrm{q: p}$
4
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In a double slit experiment, the distance between slits is increased 10 times, whereas their distance from screen is halved, the fringe width

A
remain the same.
B
becomes $\frac{1}{10}$ times.
C
becomes $\frac{1}{20}$ times.
D
becomes $\frac{1}{90}$ times.
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