Question 379: Waves Optics

Question

In Young's double slit experiment, the intensity at a point where the path difference is $$\frac{\lambda}{4}$$ [ $$\lambda$$ is wavelength of light used] is '$$\mathrm{I}$$'. If '$$\mathrm{I}_0$$' is the maximum intensity then $$\frac{\mathrm{I}}{\mathrm{I}_0}$$ is equal to $$\left[\cos \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]$$

Accepted Answer

If $$\mathrm{I}^{\prime}$$ is the intensity of each wave, then resultant intensity is given by
$$I=4 I^{\prime} \cos ^2 \frac{\phi}{2}$$
I will have maximum value when $$\cos ^2 \frac{\phi}{2}=1$$
$$	herefore$$ maximum intensity, $$\mathrm{I}_0=4 \mathrm{I}^{\prime}$$
When path difference is $$\frac{\lambda}{4}$$, the phase difference, $$\phi=\frac{\pi}{2}$$
The resultant intensity, $$I=4 I^{\prime} \cos ^2 \frac{\pi}{4} \quad=4 I^{\prime} 	imes \frac{1}{2}=2 I^{\prime}$$
$$	herefore \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{1}{2}$$

Answer

3 : 2

Answer

2 : 3

Answer

3 : 4

Answer

1 : 2

Waves Optics

Choose the correct answer: