Question

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

• A. $K$
• B. $\frac{K}{4}$
• C. $\frac{K}{2}$
• D. Zero

Answer 1

Answer: (C)

Intensity at any point on the screen is
$I=4 I_{0} \cos ^{2} \frac{\phi}{2}$
Where $I_{0}$ is the intensity of either wave
and $\phi$ is the phase difference between two
waves. Phase difference, $\phi=\frac{2 \pi}{\lambda} \times$ path difference When path difference is $\lambda$, then
$\phi=\frac{2 \pi}{\lambda} \times \lambda=2 \pi$
$\therefore I=4 I_{0} \cos ^{2}\left(\frac{2 \pi}{2}\right)$
$=4 I_{0} \cos ^{2}(\pi)=4 I_{0}=K \ldots . \rightarrow(1)$
When path difference is $\frac{\lambda}{4}$ then
$\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$
$\therefore I=4 I_{0} \cos ^{2}\left(\frac{\pi}{4}\right)$
$=2 I_{0}=\frac{K}{2}$ [ Using (1)]