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. $\frac{1}{2}$
• C. $I$
• D. zero

Answer 1

Answer: (B)

For path difference $\lambda$, phase
difference $=2 \pi\left( Q =\frac{2 \pi}{\lambda} x =\frac{2 \pi}{\lambda} \cdot \lambda=2 \pi\right)$
$\Rightarrow I = I _0+ I _0+2 I _0 \cos 2 \pi $
$\Rightarrow I =4 I _0$
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 $I^{\prime}=2 I_0=2 \cdot \frac{I}{4}=\frac{I}{2}$