Question 90: Wave Optics

Question

In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is $$7 \lambda / 4$$. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :

Accepted Answer

$$\begin{aligned}
& \Delta x=\frac{7 \lambda}{4} \
& \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{\lambda} 	imes \frac{7 \lambda}{4}=\frac{7 \pi}{2} \
& \mathrm{I}=\mathrm{I}_{\max } \cos ^2\left(\frac{\phi}{2}ight) \
& \frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^2\left(\frac{\phi}{2}ight)=\cos ^2\left(\frac{7 \pi}{2 	imes 2}ight)=\cos ^2\left(\frac{7 \pi}{4}ight) \
& =\cos ^2\left(2 \pi-\frac{\pi}{4}ight) \
& =\cos ^2 \frac{\pi}{4} \
& =\frac{1}{2} \
&
\end{aligned}$$

Answer

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

Answer

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

Answer

$$\frac{1}{3}$$

Answer

$$\frac{1}{4}$$

Wave Optics

In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is $7 \lambda / 4$. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :

Choose the correct answer: