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Wave Optics Part 1

Section: Physics

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Question : 36 of 100

Marks: +1, -0

In Young's double slit arrangement, slits are separated by a gap of

0.5\text{ mm}

, and the screen is placed at a distance of

0.5\text{ m}

from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of

5890\text{ Å}

[18 Mar 2021 Shift 1]

$1178\times10^{-9}\text{ m}$
$1178\times10^{-6}\text{ m}$
$1178\times10^{-12}\text{ m}$
$5890\times10^{-7}\text{ m}$

Solution:

Given,

The distance of sources from the screen,

D=0.5\text{ m}

The distance between the slits,

d=0.5\text{ mm}=0.0005\text{ m}

The wavelength of the monochromatic light,

\lambda=5890\text{ Å}=5890\times10^{-10}\text{ m}

Y=2\beta

As we know,

\beta=\;\frac{\lambda D}{d}

Substituting the values in the above equation.

Y=2\left(\;\frac{\lambda D}{d}\right)=2\left(\;\frac{5890\times10^{-10}\times0.5}{0.0005}\right)

=1178\times10^{-6}\text{ m}

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