Question

In a Young’s double slit interference experiment, one slit happens to have its width 4 times that of another. 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
(1) 9
(2) $\dfrac{5}{3}$
(3) 3
(4) $\dfrac{{25}}{9}$

Answer 1

Hint
The above problem is based on the concept of the interference. The interference is the phenomenon in which different coherent sources of the light produce a light pattern which is obtained by the combined effect of both coherent sources.

Complete step by step answer
Let the width of the narrow slit is then the width of the wider slit becomes 4a. The intensity of the maximum light is ${I_{\max }}$ and intensity of the minimum light is ${I_{\min }}$.
The maximum slit width occurs due to the addition of the width of narrow and wide slit. The minimum slit width occurs due to the subtraction of the wide slit and narrow slit.
The equation to find the maximum slit width of the light is given as:
$\Rightarrow {a_{\max }} = {w_n} + {w_w}......\left( 1 \right)$
Here, ${w_n}$ is the width of the narrow slit and its value is a, ${w_w}$ is the width of the wide slit and its value is 4a.
Substitute a for ${w_n}$ and 4a for ${w_w}$in the equation (1) to find the maximum intensity of the light.
$\Rightarrow {a_{\max }} = a + 4a$
$\Rightarrow {a_{\max }} = 5a$
The equation to find the minimum slit width of the light is given as:
$\Rightarrow {a_{\min }} = {w_w} - {w_n}......\left( 2 \right)$
Substitute a for ${w_n}$ and 4a for ${w_w}$in the equation (2) to find the minimum intensity of the light.
$\Rightarrow {a_{\min }} = 4a - a$
$\Rightarrow {a_{\min }} = 3a$
The intensity of the coming light is proportional to slit width, so the maximum intensity becomes proportional to the square of the maximum slit width and so the minimum intensity becomes proportional to the square of the minimum slit width.
The equation to find the ratio of the maximum to minimum intensity in the interference pattern is given as:
$\Rightarrow \dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{a_{\max }^2}}{{a_{\min }^2}}......\left( 3 \right)$
Substitute 5a for ${a_{\max }}$ and 3a for ${a_{\min }}$ in the expression (3) to calculate the ratio of the maximum to minimum intensity.
$\Rightarrow \dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{{{\left( {5a} \right)}^2}}}{{{{\left( {3a} \right)}^2}}}$
$\Rightarrow \dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{25{a^2}}}{{9{a^2}}}$
$\Rightarrow \dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{25}}{9}$
Thus, the ratio of the maximum to minimum intensity in the interference pattern is $\dfrac{{25}}{9}$ and the option (4) is the correct answer.

Note
The maximum slit width is obtained by the addition of the width of narrow slit and wide slit. The minimum slit width is obtained by the subtraction of the width of narrow slit and wide.