JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 9)
In a Young's double slit experiment, the ratio of the slit's width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be :
25 : 9
4 : 1
$${\left( {\sqrt 3 + 1} \right)^4}:16$$
9 : 1
Explanation
$${I_1} = 4{I_0}$$
$${I_2} = {I_0}$$
$${I_{\max }} = {\left( {\sqrt {{I_0}} + \sqrt {{I_2}} } \right)^2}$$
$$ = {\left( {2\sqrt {{I_0}} + \sqrt {{I_0}} } \right)^2} = 9{I_0}$$
$${I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}$$
$$ = {\left( {2\sqrt {{I_0}} - \sqrt {{I_0}} } \right)^2} = {I_0}$$
$$ \therefore $$ $${{{{\mathop{\rm I}\nolimits} _{\max }}} \over {{I_{\min }}}} = {{9{I_0}} \over {{I_0}}} = {9 \over 1}$$
$${I_2} = {I_0}$$
$${I_{\max }} = {\left( {\sqrt {{I_0}} + \sqrt {{I_2}} } \right)^2}$$
$$ = {\left( {2\sqrt {{I_0}} + \sqrt {{I_0}} } \right)^2} = 9{I_0}$$
$${I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}$$
$$ = {\left( {2\sqrt {{I_0}} - \sqrt {{I_0}} } \right)^2} = {I_0}$$
$$ \therefore $$ $${{{{\mathop{\rm I}\nolimits} _{\max }}} \over {{I_{\min }}}} = {{9{I_0}} \over {{I_0}}} = {9 \over 1}$$
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