JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 8)
Two coherent light sources having intensity in the ratio 2x produce an interference pattern. The ratio $${{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}}$$ will be :
$${{2\sqrt {2x} } \over {2x + 1}}$$
$${{2\sqrt {2x} } \over {x + 1}}$$
$${{\sqrt {2x} } \over {x + 1}}$$
$${{\sqrt {2x} } \over {2x + 1}}$$
Explanation
Given, $${{{I_1}} \over {{I_2}}} = 2x$$
We know,
$${{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}$$
$$ = {\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}$$
$$ = {\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)^2}$$
Now,
$${{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}}$$
$$ = {{{{{I_{\max }}} \over {{I_{\min }}}} - 1} \over {{{{I_{\max }}} \over {{I_{\min }}}} + 1}}$$
$$ = {{{{\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)}^2} - 1} \over {{{\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)}^2} + 1}}$$
$$ = {{{{\left( {\sqrt {2x} + 1} \right)}^2} - {{\left( {\sqrt {2x} - 1} \right)}^2}} \over {{{\left( {\sqrt {2x} + 1} \right)}^2} + {{\left( {\sqrt {2x} - 1} \right)}^2}}}$$
$$ = {{4\sqrt {2x} } \over {2 + 4x}} = {{2\sqrt {2x} } \over {1 + 2x}}$$
We know,
$${{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}$$
$$ = {\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}$$
$$ = {\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)^2}$$
Now,
$${{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}}$$
$$ = {{{{{I_{\max }}} \over {{I_{\min }}}} - 1} \over {{{{I_{\max }}} \over {{I_{\min }}}} + 1}}$$
$$ = {{{{\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)}^2} - 1} \over {{{\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)}^2} + 1}}$$
$$ = {{{{\left( {\sqrt {2x} + 1} \right)}^2} - {{\left( {\sqrt {2x} - 1} \right)}^2}} \over {{{\left( {\sqrt {2x} + 1} \right)}^2} + {{\left( {\sqrt {2x} - 1} \right)}^2}}}$$
$$ = {{4\sqrt {2x} } \over {2 + 4x}} = {{2\sqrt {2x} } \over {1 + 2x}}$$
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