In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :

[NEET 2020]

(a) double (b) half (c) four times (d) one-fourth

Given:

$d_2 = \frac{d_1}{2}$

and,

$D_2 = 2 D_1$

The Fringe width is given by

$\beta _1 =\frac{\lambda D_1}{d_1}$

- - - - (1)

The new fringe width will be

$\beta_2 =\frac{\lambda D_2}{d_2}$

$\beta_2 =\frac{\lambda (2D_1)}{\frac{(d_1)}{2}}$

$\beta_2 =\frac{\lambda (4D_1)}{(d_1)}$

From equation no (1)

$\beta_2 =4\beta_1$

i.e. New fringe width becomes four times the initial fringe width

Therefore, the correct option is option 'C'

Milan Tomic

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