JEE Main 2017 — Physics Optics

2017mcqmedium

In a Young's double slit experiment, slits are separated by $0.5\mathrm{mm}$ , and the screen is placed $150\mathrm{cm}$ away. A beam of light consisting of two wavelengths, $650\mathrm{nm}$ and $520\mathrm{nm}$ , is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

Official previous-year question

Held on 2 Apr 2017 · Verified 6 Jul 2026.

Options

A
$15.6\mathrm{mm}$
B
$1.56\mathrm{mm}$
C
$7.8\mathrm{mm}$
D
$9.75\mathrm{mm}$

Did you get this right?

Solution

Given,

Wavelength of first light is, ${\lambda }_{1}=650\mathrm{nm}$

Wavelength of second light is, ${\lambda }_{2}=520\mathrm{nm}$

Screen distance is, $D=150\mathrm{cm}$

Slit width is, $d=0.5\mathrm{mm}$

For first wavelength, location of fringe from central line,

${y}_{1}=\frac{{n}_{1}{\lambda }_{1}D}{d}$ for bright fringes

For second wavelength, location of fringe from central line,

${y}_{2}=\frac{{n}_{2}{\lambda }_{2}D}{d}$ for bright fringes

Condition of coincide of fringe is,

${y}_{1}={y}_{2}$

$\Rightarrow {n}_{1}{\lambda }_{1}={n}_{2}{\lambda }_{2}$

$\Rightarrow {n}_{1}\times 650={n}_{2}\times 520$

$\Rightarrow \frac{{n}_{1}}{{n}_{2}}=\frac{4}{5}$

For the minimum value of ${n}_{1}$ and ${n}_{2}$

${n}_{1}=4$ and ${n}_{2}=5$

${({y}_{1})}_{min}=4\times 650\mathrm{nm}\times \frac{1.50m}{0.5\times {10}^{-3}m}$

$=7800\times {10}^{-6}m$

$=7.8 \mathrm{mm}$

Your note

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Solution

Given,

Wavelength of first light is, ${\lambda }_{1}=650\mathrm{nm}$

Wavelength of second light is, ${\lambda }_{2}=520\mathrm{nm}$

Screen distance is, $D=150\mathrm{cm}$

Slit width is, $d=0.5\mathrm{mm}$

For first wavelength, location of fringe from central line,

${y}_{1}=\frac{{n}_{1}{\lambda }_{1}D}{d}$ for bright fringes

For second wavelength, location of fringe from central line,

${y}_{2}=\frac{{n}_{2}{\lambda }_{2}D}{d}$ for bright fringes

Condition of coincide of fringe is,

${y}_{1}={y}_{2}$

$\Rightarrow {n}_{1}{\lambda }_{1}={n}_{2}{\lambda }_{2}$

$\Rightarrow {n}_{1}\times 650={n}_{2}\times 520$

$\Rightarrow \frac{{n}_{1}}{{n}_{2}}=\frac{4}{5}$

For the minimum value of ${n}_{1}$ and ${n}_{2}$

${n}_{1}=4$ and ${n}_{2}=5$

${({y}_{1})}_{min}=4\times 650\mathrm{nm}\times \frac{1.50m}{0.5\times {10}^{-3}m}$

$=7800\times {10}^{-6}m$

$=7.8 \mathrm{mm}$

JEE Main 2017 — Physics Optics

Options

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Solution

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JEE Main 2017 — Physics Optics

Options

Did you get this right?

Solution

Your note

JEE Main Optics in other years

More JEE Main Optics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics