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 :
Solution
Given, ${{{I_1}} \over {{I_2}}} = 2x$<br><br>We know,<br><br>$${{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}$$<br><br>$$ = {\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}$$<br><br>$= {\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)^2}$<br><br>Now, <br><br>${{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}}$<br><br>$$ = {{{{{I_{\max }}} \over {{I_{\min }}}} - 1} \over {{{{I_{\max }}} \over {{I_{\min }}}} + 1}}$$<br><br>$$ = {{{{\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)}^2} - 1} \over {{{\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)}^2} + 1}}$$<br><br>$$ = {{{{\left( {\sqrt {2x} + 1} \right)}^2} - {{\left( {\sqrt {2x} - 1} \right)}^2}} \over {{{\left( {\sqrt {2x} + 1} \right)}^2} + {{\left( {\sqrt {2x} - 1} \right)}^2}}}$$<br><br>$= {{4\sqrt {2x} } \over {2 + 4x}} = {{2\sqrt {2x} } \over {1 + 2x}}$
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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