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In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is K, ( $\lambda$ being the wave length of light used). The intensity at a point where the path difference is $\frac{\lambda }{4}$ , will be:

Option 1)
$K$

Option 2)
$\frac{K}{4}$

Option 3)
$\frac{K}{2}$

Option 4)
$Zero$

Answers (1)

Resultant Intensity of two wave -

$I= I_{1}+I_{2}+2sqrt{I_{1}I_{2}}cos heta$

- wherein

$I_{1}=$ Intencity of wave 1

$I_{2}=$ Intencity of wave 2

$heta =$ Phase difference

at a point where path difference is $\lambda=>phase difference =2\pi$

$\therefore I_{R}=K-(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}$

$\therefore I_{R}=K-(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}$ $=>K=4 I-----------------(1)$

$(I_{1} = I_{2}=I)$

When path a difference is $\frac{\lambda}{4}$

$=>$ phase difference = $\frac{\pi}{2}$

$I_{R}=I_{1} + I_{2} = 2I=\frac{k}{2}$

Option 1)

$K$

This option is incorrect

Option 2)

$\frac{K}{4}$

This option is incorrect

Option 3)

$\frac{K}{2}$

This option is correct

Option 4)

$Zero$

This option is incorrect

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subam

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