Electromagnetism
Current Electricity
MCQ (Single Correct Answer)
Moving Charges and Magnetism
MCQ (Single Correct Answer)
Magnetism and Matter
MCQ (Single Correct Answer)
Electromagnetic Waves
MCQ (Single Correct Answer)
Electromagnetic Induction
MCQ (Single Correct Answer)
Alternating Current
MCQ (Single Correct Answer)
Modern Physics
Dual Nature of Radiation
MCQ (Single Correct Answer)
Semiconductor Devices and Logic Gates
MCQ (Single Correct Answer)
Communication Systems
MCQ (Single Correct Answer)
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
$\sin ^{-1}\left(\frac{\lambda}{4 d}\right)$
Questions Asked from MCQ (Single Correct Answer)
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