1
MHT CET 2026 20th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
To improve the resolving power of a compound microscope we should
A
decrease the diameter of the objective lens.
B
decrease the focal length of the eye-piece.
C
increase the refractive index of the medium between the object and objective lens.
D
increase the wavelength of light.
2
MHT CET 2026 20th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
In the diffraction pattern, the first maximum is at $30^\circ$, when a monochromatic light of wavelength $\lambda$ is incident on a slit of width $a$. For the same wavelength, if the slit width is changed, so that first maximum is at $45^\circ$. The slit width is changed by $\left(\sin 30^\circ = \dfrac{1}{2}, \sin 45^\circ = \dfrac{1}{\sqrt{2}}\right)$
A
$3\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)\lambda$
B
$\dfrac{3}{\sqrt{2}}\lambda$
C
$3\sqrt{2}\lambda$
D
$3\left(\dfrac{\sqrt{2}+1}{\sqrt{2}}\right)\lambda$
3
MHT CET 2026 20th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
Light of wavelength $580$ nm is incident normally on a slit of width '$a$'. The distance between slit and screen is $2.5$ m and the distance of the second order maximum from the centre of the screen is $14.5$ mm in a diffraction pattern. The value of '$a$' is
A
$0.12 \times 10^{-3}\,\text{m}$
B
$0.25 \times 10^{-3}\,\text{m}$
C
$0.36 \times 10^{-3}\,\text{m}$
D
$0.50 \times 10^{-3}\,\text{m}$
4
MHT CET 2026 20th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$ . The ratio $\dfrac{\lambda_2}{\lambda_1}$ is
A
$\dfrac{m}{n-1}$
B
$\dfrac{m}{(2n-1)}$
C
$\dfrac{2m}{(2n-1)}$
D
$\dfrac{2m}{(2n+1)}$

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