JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 17)
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 :
$$\frac{1}{2}$$
$$\frac{3}{4}$$
$$\frac{1}{3}$$
$$\frac{1}{4}$$
Explanation
$$\begin{aligned}
& \Delta x=\frac{7 \lambda}{4} \\
& \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{\lambda} \times \frac{7 \lambda}{4}=\frac{7 \pi}{2} \\
& \mathrm{I}=\mathrm{I}_{\max } \cos ^2\left(\frac{\phi}{2}\right) \\
& \frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^2\left(\frac{\phi}{2}\right)=\cos ^2\left(\frac{7 \pi}{2 \times 2}\right)=\cos ^2\left(\frac{7 \pi}{4}\right) \\
& =\cos ^2\left(2 \pi-\frac{\pi}{4}\right) \\
& =\cos ^2 \frac{\pi}{4} \\
& =\frac{1}{2} \\
&
\end{aligned}$$
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