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)
S subam

 

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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