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Let $S_{1}\; and\; S_{2}$ be the two slits in Young’s double slit experiment. If central maxima is observed at P and angle $\angle S_{1}PS_{2}=\theta ,$ then the fringe width for the light of wavelength $\lambda$ will be. (Assume $\theta$ to be a small angle)
Option: 1
$\frac{\lambda }{\theta }$

Option: 2
$\lambda \theta$

Option: 3
$\frac{2\lambda }{\theta }$

Option: 4
$\frac{\lambda }{2\theta }$

Answers (1)

best_answer

Fringe Width -

$\beta = \frac{\lambda D}{d}$

- wherein

$\beta = y_{n+1}-y_{n}$

$y_{n+1}=$ Distance of $\left ( n+1 \right )^{th}$

Maxima $= \left ( n+1 \right )\frac{\lambda D}{d}$

$y_{n}=$ Distance of $n^{th}$

maxima $= \frac{n\lambda D}{d}$

In $\Delta S_{1}PO$

$tan\theta =\frac{\theta }{2}=\frac{d/2}{D}$

As $D> > d$

\ $\theta$ is very small.

$\Rightarrow tan \frac{\theta }{2}\approx \frac{\theta }{2}$

$\Rightarrow \frac{\theta }{2}=\frac{d}{2D}\Rightarrow \frac{D}{d}=\frac{1}{\theta }\Rightarrow\; Fringe\; width=\frac{\lambda D}{d}=\frac{\lambda }{\theta }$

Posted by

sudhir.kumar

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