Get Answers to all your Questions

header-bg qa

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)


  • Option 2)


  • Option 3)


  • Option 4)



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)


This option is incorrect 

Option 2)


This option is incorrect 

Option 3)


This option is correct 

Option 4)


This option is incorrect 

Posted by


View full answer

Crack NEET with "AI Coach"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support