Question 347: Waves Optics

Question

In Young's double slit experiment, the wavelength of light used is '$$\lambda$$'. The intensity at a point is '$$\mathrm{I}$$' where path difference is $$\left(\frac{\lambda}{4}\right)$$. If $$I_0$$ denotes the maximum intensity, then the ratio $$\left(\frac{\mathrm{I}}{\mathrm{I}_0}\right)$$ is
$$\left(\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right)$$

Accepted Answer

$$\begin{aligned}
& 	ext { Phase difference, } \Delta \phi=\left(\frac{2 \pi}{\lambda}ight) \Delta l \
& 	ext { For path difference } \frac{\lambda}{4}, \
& 	ext { Phase difference } \Delta \phi=\frac{\pi}{2} \
& 	ext { Using, } \mathrm{I}=\mathrm{I}_0 \cos ^2 \frac{\phi}{2} \
& 	herefore \quad \frac{\mathrm{I}}{\mathrm{I}_0}=\cos ^2 \frac{\phi}{2}=\cos ^2\left(\frac{\pi}{4}ight) \
& 	herefore \quad \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{1}{2}
\end{aligned}$$

Answer

$$\frac{1}{\sqrt{2}}$$

Answer

$$\frac{1}{2}$$

Answer

$$\frac{3}{4}$$

Answer

$$\frac{\sqrt{3}}{2}$$

Waves Optics

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