JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 9)
In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
4 : 1
2 : 1
1 : 4
3 : 1
Explanation
Given, amplitude $$\propto$$ width of slit
$$\Rightarrow$$ A2 = 3A1
We know that,
$${{{I_{\max }}} \over {{I_{\min }}}} = {{{{(\sqrt {{I_1}} + \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {{I_1}} - \sqrt {{I_2}} )}^2}}}$$
$$\because$$ Intensity, I $$\propto$$ A2
$$\therefore$$ $${{{I_{\max }}} \over {{I_{\min }}}} = {{{{({A_1} + {A_2})}^2}} \over {{{(|{A_1} - {A_2}|)}^2}}}$$
$$ = {\left( {{{{A_1} + 3{A_1}} \over {|{A_1} - 3{A_1}|)}}} \right)^2} = {\left( {{{4{A_1}} \over {2{A_1}}}} \right)^2} = {4 \over 1}$$
$$\therefore$$ $${I_{\max }}:{I_{\min }} = 4:1$$
$$\Rightarrow$$ A2 = 3A1
We know that,
$${{{I_{\max }}} \over {{I_{\min }}}} = {{{{(\sqrt {{I_1}} + \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {{I_1}} - \sqrt {{I_2}} )}^2}}}$$
$$\because$$ Intensity, I $$\propto$$ A2
$$\therefore$$ $${{{I_{\max }}} \over {{I_{\min }}}} = {{{{({A_1} + {A_2})}^2}} \over {{{(|{A_1} - {A_2}|)}^2}}}$$
$$ = {\left( {{{{A_1} + 3{A_1}} \over {|{A_1} - 3{A_1}|)}}} \right)^2} = {\left( {{{4{A_1}} \over {2{A_1}}}} \right)^2} = {4 \over 1}$$
$$\therefore$$ $${I_{\max }}:{I_{\min }} = 4:1$$
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