1
MHT CET 2026 11th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
Two polaroid A and B placed in such a way that the pass-axis of polaroid are perpendicular to each other. Now, another polaroid C is placed between A and B bisecting angle between them. If intensity of unpolarised light is $I_0$ then intensity of transmitted light after passing through polaroid B will be
A
$\dfrac{I_0}{4}$
B
$\dfrac{I_0}{2}$
C
$\dfrac{I_0}{8}$
D
zero
2
MHT CET 2026 11th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
In a Young's double slit experiment, the intensities at two points, for the path difference $\dfrac{\lambda}{4}$ and $\dfrac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits, then $\dfrac{I_1 + I_2}{I_0} = $
$\left(\cos 45^\circ = \dfrac{1}{\sqrt{2}}, \cos 60 = \dfrac{1}{2}\right)$
A
$2$
B
$3$
C
$4$
D
$5$
3
MHT CET 2026 11th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
Two light waves amplitudes in the ratio $3:1$ produce interference. The ratio of the maximum to minimum intensity is
A
$9 : 4$
B
$16 : 9$
C
$3 : 2$
D
$4 : 1$
4
MHT CET 2026 11th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays ($v_1 > v_2$). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $k : 1$, then what is the threshold frequency of the metallic surface?
A
$\dfrac{v_1 - v_2}{k}$
B
$\dfrac{v_1 - v_2}{k - 1}$
C
$\dfrac{kv_1 - v_2}{k - 1}$
D
$\dfrac{kv_2 - v_1}{k - 1}$

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