NEET UG 2016 — Physics Optics

2016mcqhard

The intensity at the maximum in a Young's double slit experiment is ${I}_{0}$ . Distance between two slits is $d=5\lambda$ , where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D=10d$ ?

Official previous-year question

Held on 30 Apr 2016 · Verified 9 Jul 2026.

Options

A
${I}_{0}$
B
$\frac{{I}_{0}}{4}$
C
$\frac{3}{4}{I}_{0}$
D
$\frac{{I}_{0}}{2}$

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Solution

In YDSE ${I}_{max}={I}_{0}$

Path difference at a point in front of one of shifts is

$\Delta x=d(\frac{y}{D})=d(\frac{\frac{d}{2}}{D})=\frac{{d}^{2}}{2D}$ $( Here y=\frac{d}{2} )$

$\Delta x=\frac{{d}^{2}}{2(10d)}=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}$

Path difference is

$\phi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }(\frac{\lambda }{4})$

$\phi =\frac{\pi }{2}$

So intensity at that point is

$I={I}_{max}{\mathrm{cos}}^{2}(\frac{\theta }{2})$

$I={I}_{0}{\mathrm{cos}}^{2}(\frac{\pi }{4})=\frac{{I}_{0}}{2}$

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Solution

In YDSE ${I}_{max}={I}_{0}$

Path difference at a point in front of one of shifts is

$\Delta x=d(\frac{y}{D})=d(\frac{\frac{d}{2}}{D})=\frac{{d}^{2}}{2D}$ $( Here y=\frac{d}{2} )$

$\Delta x=\frac{{d}^{2}}{2(10d)}=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}$

Path difference is

$\phi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }(\frac{\lambda }{4})$

$\phi =\frac{\pi }{2}$

So intensity at that point is

$I={I}_{max}{\mathrm{cos}}^{2}(\frac{\theta }{2})$

$I={I}_{0}{\mathrm{cos}}^{2}(\frac{\pi }{4})=\frac{{I}_{0}}{2}$

NEET UG 2016 — Physics Optics

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Solution

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NEET UG 2016 — Physics Optics

Options

Did you get this right?

Solution

Your note

NEET UG Optics in other years

More NEET UG Optics questions

Other Physics topics for NEET UG

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics