Question

The slits in a Young's double slit experiment have equal widths and the source is placed symmetrically relative to the slits. The intensity at the central fringes is $I_0$. If one of the slits is closed, the intensity at this point will now be :-

Answer 1

CASE-1 :

Resultant amplitude , $A_{max}=A_1+A_2$

$A_1=A+A=2A$

$\because$ Intensity , $I\propto (\text{Amplitude}{)}^2$

we get, $I\propto (2A{)}^2\text{or}{I}_0=4K{A}^2\dots .(i)$

CASE-2:

When one of the slits is closed , then the resultant amplitude is, $A_2=A+0=A$

Intensity is, $I=K{A}^2\dots .\left(ii\right)$

From equations (i) and (ii)

$\frac{I_0}{I}=\frac{4K{A}^2}{K{A}^2}=4$ or $I=\frac{I_0}{4}$

Hence, option (b) is the correct answer.