1
MHT CET 2026 16th April Evening 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 $P_3$ are inclined at angle of $60^\circ$ and $90^\circ$ with respect to axis of $P_1$. The source has an intensity $I_0$. The intensity of light finally coming out is
A
$\dfrac{I_0}{2}\cos^2 30^\circ\cos^2 90^\circ$
B
$\dfrac{I_0}{2}\cos^2 60^\circ\cos^2 30^\circ$
C
$I_0\cos^2 60^\circ\cos^2 30^\circ$
D
zero
2
MHT CET 2026 16th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
In Young's double slit experiment, two slits are illuminated with a light of wavelength $\lambda$. The line joining $A_1P$ is perpendicular to $A_1A_2$ as shown in figure. If the first minimum is detected at P, the value of slits separation 'a' will be
(D = distance between source and screen)
MHT CET 2026 16th April Evening Shift Physics - Wave Optics Question 8 English
A
$\lambda\text{D}$
B
$\sqrt{\lambda\text{D}}$
C
$\sqrt{\dfrac{\lambda}{\text{D}}}$
D
$\sqrt{\dfrac{\text{D}}{\lambda}}$
3
MHT CET 2026 16th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
When a light of wavelength '$\lambda$' falls on the emitter of a photosensitive surface, maximum speed of emitted photoelectrons is 'V'. If the incident wavelength is changed to '$2\lambda/3$', maximum speed of emitted photoelectrons will be
A
less than $V(1.5)^{\frac{1}{2}}$
B
greater than $V(1.5)^{\frac{1}{2}}$
C
less than $V$
D
less than $\dfrac{V}{2}$
4
MHT CET 2026 16th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
A photoemissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectron is $\lambda_0$. Expression for de-Broglie wavelength is (m = mass of electron, h = Planck's constant, C = Speed of light)
A
$(\text{h}\lambda_i/2\text{mc})^{\frac{1}{2}}$
B
$(\text{h}\lambda_0/2\text{mc})^{\frac{1}{2}}$
C
$[\text{h}/2\text{mc}(\dfrac{1}{\lambda_i} - \dfrac{1}{\lambda_0})]^{\frac{1}{2}}$
D
$[\text{h}/[2\text{mc}(\dfrac{1}{\lambda_i} - \dfrac{1}{\lambda_0})]^{\frac{1}{2}}$

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