Wave Optics - Result Question 15
15. 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 $\lambda / 4$, will be: [2014]
(a) $K$
(b) $K / 4$
(c) $K / 2$
(d) Zero
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Answer:
Correct Answer: 15. (c)
Solution:
- (c) For path difference $\lambda$, phase difference $=2 \pi rad$.
For path difference $\frac{\lambda}{4}$, phase difference $=\frac{\pi}{2} rad$.
As $K=4 I_0$ so intensity at given point where path difference is $\frac{\lambda}{4}$
$K^{\prime}=4 I_0 \cos ^{2}(\frac{\pi}{4})(\cos \frac{\pi}{4}=\cos 45^{\circ})=2 I_0=\frac{K}{2}$
