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Wave Optics Part 1

Section: Physics

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Question : 93 of 100

Marks: +1, -0

In a Young's double slit experiment the intensity at a point where the path difference is

\;\frac{\lambda}{6}(\lambda

being the wavelength of light used ) is

I

I_0

denotes the maximum intensity,

\;\frac{I}{I_0}

$\frac{3}{4}$
$\frac{1}{\sqrt{2}}$
$\frac{\sqrt{3}}{2}$
$\frac{1}{2}$

Solution:

The intensity of light at any point of the screen where the phase difference due to light coming from the two slits is

\varphi

is given by

I=I_0 \cos ^2 \left(\frac{\varphi}{2}\right)

where

I_0

is the maximum intensity.

NOTE : This formula is applicable when

I_1=I_2

.

Here

\varphi = \frac{\pi}{3}

\therefore \;\frac{I}{I_0}=\cos ^2 \;\frac{\pi}{6}= \left(\frac{\sqrt{3}}{2}\right)^2=\;\frac{3}{4}

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