First slide

Young's double slit Experiment

Question

In a Young’s double slit experiment, D equals the distance of screen from the slits and d is the separation between the slits. The distance of the nearest point from the central maximum where the intensity is same as that due to a single slit, is equal to

Moderate

A

$\frac{λD}{d}$
B

$\frac{λD}{2 d}$
C

$\frac{λD}{3 d}$
D

$\frac{2 λD}{d}$

Solution

$I = 4 I_{0} \cos^{2} (\frac{πd}{λD}) y$
Where y is the distance of a point on the screen measured from central bright fringe.
$Where I = I_{0}, \cos^{2} (\frac{πd}{λD}) y = \frac{1}{4}$
$\Rightarrow \cos (\frac{πd}{λD}) y = \pm \frac{1}{2}$
$\Rightarrow (\frac{πd}{λD}) y = \frac{π}{3}, (2 π - \frac{π}{3}), (2 π + \frac{π}{3}) etc$
$∴ y_{\min} = \frac{λD}{3 d}$

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