JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 16)
Find the ratio of maximum intensity to the minimum intensity in the interference pattern if the widths of the two slits in Young's experiment are in the ratio of 9 : 16. (Assuming intensity of light is directly proportional to the width of slits)
3 : 4
4 : 3
7 : 1
49 : 1
Explanation
Let,
Width of first slit = w1 and width of second slit = w2
Given, $${{{w_1}} \over {{w_2}}} = {9 \over {16}}$$
Given,
Intensity of light $$(I) \propto w$$
$$\therefore$$ $${{{I_1}} \over {{I_2}}} = {{{w_1}} \over {{w_2}}} = {9 \over {16}}$$
We know,
$${{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}$$
$$ = {\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}$$
$$ = {\left( {{{{3 \over 4} + 1} \over {{3 \over 4} - 1}}} \right)^2}$$
$$ = {\left( {{7 \over 1}} \right)^2}$$
$$ = {{49} \over 1}$$
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