1
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
$\mathrm{q: p}$
2
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.
3
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The angular separation of the central maximum in the Fraunhofer diffraction pattern is measured. The slit is illuminated by the light of wavelength $6000 \mathop A\limits^o$. If the slit is illuminated by light of another wavelength, the angular separation decreases by $20 \%$. The wavelength of light used is

A
$6400 \mathop A\limits^o$
B
$5600 \mathop A\limits^o$
C
$4800 \mathop A\limits^o$
D
$4400 \mathop A\limits^o$
4
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

In Young's double slit experiment, intensity at a point is $\left(\frac{1}{4}\right)$ of the maximum intensity. The angular position of this point is

A
$\sin ^{-1}\left(\frac{\lambda}{D}\right)$
B
$\sin ^{-1}\left(\frac{\lambda}{2 d}\right)$
C
$\sin ^{-1}\left(\frac{\lambda}{3 \mathrm{~d}}\right)$
D
$\sin ^{-1}\left(\frac{\lambda}{4 d}\right)$
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