# 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) $Zero$

S subam

Resultant Intensity of two wave -

$I= I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos \theta$

- wherein

$I_{1}=$ Intencity of wave 1

$I_{2}=$ Intencity of wave 2

$\theta =$ 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

Exams
Articles
Questions