Question 405: Wave Optics

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 :

Accepted Answer

Given that the slit distance is made half and the screen distance made double than the original value, then,
Fringe width, $$\beta  = {{\lambda D} \over d}$$
Now, $$d' = {d \over 2}$$
 and $$D' = 2D$$
So, $$\beta ' = {{\lambda (2D)} \over {d/2}} = {{4\lambda D} \over d}$$
$$\beta $$ = 4$$\beta $$

Answer

half

Answer

four times

Answer

one-fourth

Answer

double

Wave Optics

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 :

Choose the correct answer: