Question

In young's double-slit experiment the intensity of light at a point on the screen where the path difference is $\lambda$ is $I$, $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{\lambda }{4}$ will be

• A. $\frac{I}{4}$
• B. $\frac{I}{2}$
• C. $I$
• D. zero

Answer 1

Answer: (B)

$\frac{I}{2}$

For path difference $\lambda$, phase
difference = 2$\pi$ $(Q=\frac{2\pi }{\lambda }x=\frac{2\pi }{\lambda }\lambda =2\pi )$
$\Rightarrow I_0=I_0+I_0+2I_{0}cos2\pi$
$\Rightarrow I_0=4I_0$ ($\therefore cos2\pi$ = 1 )
For x = $\frac{\lambda }{4}$, phase difference = $\frac{\pi }{2}$
$\therefore I^{\prime }$ =$I_1$ + $I_2$ + 2$\sqrt{I_1}$ $\sqrt{I_2}$ cos $\frac{\pi }{2}$
If $I_1$= $I_2$=$I_0$ then $I^{\prime }=2I_0$ = 2.$\frac{I}{4}$ = $\frac{I}{2}$