Question 416: Waves Optics

Question

The interference pattern is obtained with two coherent light sources of intensity ratio n. In  the interference pattern, the ratio $${{{I_{max}} - {I_{\min }}} \over {{I_{max}} + {I_{min}}}}$$ will be

Accepted Answer

Maximum Intensity is given as I
max
= $${\left( {\sqrt {{I_1}}  + \sqrt {{I_2}} } \right)^2}$$
Minimum intensity is given as I
min
= $${\left( {\sqrt {{I_1}}  - \sqrt {{I_2}} } \right)^2}$$
Given $${{{I_1}} \over {{I_2}}}$$ = n
$${{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}}  + \sqrt {{I_2}} } \over {\sqrt {{I_1}}  - \sqrt {{I_2}} }}} \right)^2}$$
= $${\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}}  + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}}  - 1}}} \right)^2}$$ = $${\left( {{{\sqrt n  + 1} \over {\sqrt n  - 1}}} \right)^2}$$
$${{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}} = $$ $${{{{{I_{\max }}} \over {{I_{\min }}}} - 1} \over {{{{I_{\max }}} \over {{I_{\min }}}} + 1}}$$
= $${{{{\left( {{{\sqrt n  + 1} \over {\sqrt n  - 1}}} \right)}^2} - 1} \over {{{\left( {{{\sqrt n  + 1} \over {\sqrt n  - 1}}} \right)}^2} + 1}}$$
= $${{{{\left( {\sqrt n  + 1} \right)}^2} - {{\left( {\sqrt n  - 1} \right)}^2}} \over {{{\left( {\sqrt n  + 1} \right)}^2} + {{\left( {\sqrt n  - 1} \right)}^2}}}$$
= $${{4\sqrt n } \over {2\left( {n + 1} \right)}}$$ = $${{2\sqrt n } \over {n + 1}}$$

Answer

$${{\sqrt n } \over {n + 1}}$$

Answer

$${{2\sqrt n } \over {n + 1}}$$

Answer

$${{\sqrt n } \over {{{\left( {n + 1} \right)}^2}}}$$

Answer

$${{2\sqrt n } \over {{{\left( {n + 1} \right)}^2}}}$$

Waves Optics

The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio ${{{I_{max}} - {I_{\min }}} \over {{I_{max}} + {I_{min}}}}$ will be

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