Question

In a Young’s double slit interference experiment, one slit happens to have its width 4 times that of the other. Assuming that intensity of out coming light is proportional to slit width, the ratio of the maximum to minimum intensity in the interference pattern is:
A.$9$.
B.$\dfrac{5}{3}$.
C.$3$.
D.$\dfrac{{25}}{9}$.

Answer 1

The Young’s double slit experiment is when a monochromatic light passing through two narrow slits then there is formation of dark fringes and bright fringes, these fringes is caused due to the superposition of the light waves from the two slits.

Complete answer:
It is given in the problem that in a Young’s double slit interference experiment, one slit happens to have its width 4 times that of the other if the intensity of out coming light is proportional to slit width and we need to find the ratio of the maximum to minimum intensity in the interference pattern.
The intensity of light through the larger width is equal to ${I_1}$ and the intensity of the light through the smaller width is equal to${I_2}$.
Let the intensity from the smaller slit be $I$ and let the intensity of the larger slit be $4I$.
The maximum intensity is equal to,
$\Rightarrow {I_{\max .}} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}}$
$\Rightarrow {I_{\max .}} = 4I + I + 2\sqrt {4I \times I}$
$\Rightarrow {I_{\max .}} = 5I + 2\sqrt {4{I^2}}$
$\Rightarrow {I_{\max .}} = 5I + 4I$
$\Rightarrow {I_{\max .}} = 9I$………eq. (1)
The minimum intensity is equal to,
$\Rightarrow {I_{\min .}} = {I_1} + {I_2} - 2\sqrt {{I_1}{I_2}}$
$\Rightarrow {I_{\min .}} = 4I + I - 2\sqrt {4I \times I}$
$\Rightarrow {I_{\min .}} = 5I - 2\sqrt {4{I^2}}$
$\Rightarrow {I_{\min .}} = 5I - 2 \times 2I$
$\Rightarrow {I_{\min .}} = 5I - 4I$
$\Rightarrow {I_{\min .}} = I$………eq. (2)
The ratio of ${I_{\max .}}$and ${I_{\min .}}$ is equal to,
$\Rightarrow \dfrac{{{I_{\max .}}}}{{{I_{\min .}}}}$
From equation (1) and equation (2) put the values of ${I_{\max .}}$ and ${I_{\min .}}$ in the above relation, we get.
$\Rightarrow \dfrac{{{I_{\max .}}}}{{{I_{\min .}}}} = \dfrac{{9I}}{I}$
$\Rightarrow \dfrac{{{I_{\max .}}}}{{{I_{\min .}}}} = 9$.
The ratio of maximum intensity to the minimum intensity is equal to 9.

The correct answer is option A.

Note:
The Young’s slit experiment makes different fringes due to overlapping of waves from the narrow slits. There is occurrence of central minima on the screen and also occurrence of central maxima on the screen.