MHT CET 2026 20th April Evening Shift | Wave Optics Question 1 | Physics

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 $\left(\sin 30^\circ = \dfrac{1}{2}, \sin 45^\circ = \dfrac{1}{\sqrt{2}}\right)$

$3\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)\lambda$

$\dfrac{3}{\sqrt{2}}\lambda$

$3\sqrt{2}\lambda$

$3\left(\dfrac{\sqrt{2}+1}{\sqrt{2}}\right)\lambda$

MHT CET 2026 20th April Morning Shift

MCQ (Single Correct Answer)

-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

$0.12 \times 10^{-3}\,\text{m}$

$0.25 \times 10^{-3}\,\text{m}$

$0.36 \times 10^{-3}\,\text{m}$

$0.50 \times 10^{-3}\,\text{m}$

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