JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 14)
The interference pattern is obtained with two coherent light sources of intensity ratio 4 : 1. And the ratio $${{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}}$$ is $${5 \over x}$$. Then, the value of x will be equal to :
3
4
2
1
Explanation
$${{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} = {{{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} + {I_1} + {I_2} - 2\sqrt {{I_1}{I_2}} } \over {{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} - {I_1} - {I_2} + 2\sqrt {{I_1}{I_2}} }}$$
$$ = {{2({I_1} + {I_2})} \over {4\sqrt {{I_1}{I_2}} }}$$
$$ = {{\left( {{{{I_1}} \over {{I_2}}} + 1} \right)} \over {2\sqrt {{{{I_1}} \over {{I_2}}}} }} = {{4 + 1} \over {2 \times 2}} = {5 \over 4}$$
So $$x = 4$$
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