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:

Question

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: (A) double (B) half (C) four times (D) one-fourth

Tardigrade Admin · Accepted Answer

Fringe width, β=(λ D/d) β' = (λ D'/d') Now, d'=(d/2) and D'=2 D So, β'=(λ × 2 D/d / 2) =(4 λ D/d) β'=4 β Fringe width becomes 4 times

Q. 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:

double

half

four times

one-fourth

Fringe width, $β = \frac{λ D}{d}$
$β^{'} = \frac{λ D ^{'}}{d ^{'}}$
Now, $d^{'} = \frac{d}{2}$ and
$D^{'} = 2 D$
So, $β^{'} = \frac{λ \times 2 D}{d /2}$
$= \frac{4 λ D}{d}$
$β^{'} = 4 β$
Fringe width becomes 4 times

Q. 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:

double

half

four times

one-fourth

Fringe width, β=dλD​ β′=d′λD′​ Now, d′=2d​ and D′=2D So, β′=d/2λ×2D​ =d4λD​ β′=4β Fringe width becomes 4 times

Fringe width, $β = \frac{λ D}{d}$
$β^{'} = \frac{λ D ^{'}}{d ^{'}}$
Now, $d^{'} = \frac{d}{2}$ and
$D^{'} = 2 D$
So, $β^{'} = \frac{λ \times 2 D}{d /2}$
$= \frac{4 λ D}{d}$
$β^{'} = 4 β$
Fringe width becomes 4 times