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Wave Optics question

2016 · Q133
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Wave Optics question

2016 · Q133

NEETPhysicsWave OpticsMCQ+4 / −1
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio Imax−Imin⁡Imax+Imin{{{I_{max}} - {I_{\min }}} \over {{I_{max}} + {I_{min}}}}Imax​+Imin​Imax​−Imin​​ will be
  1. A
    nn+1{{\sqrt n } \over {n + 1}}n+1n​​
  2. B
    2nn+1{{2\sqrt n } \over {n + 1}}n+12n​​
  3. C
    n(n+1)2{{\sqrt n } \over {{{\left( {n + 1} \right)}^2}}}(n+1)2n​​
  4. D
    2n(n+1)2{{2\sqrt n } \over {{{\left( {n + 1} \right)}^2}}}(n+1)22n​​
View written solutionFree

Correct answer: B

Maximum Intensity is given as Imax = (I1+I2)2{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}(I1​​+I2​​)2

Minimum intensity is given as Imin = (I1−I2)2{\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}(I1​​−I2​​)2

Given I1I2{{{I_1}} \over {{I_2}}}I2​I1​​ = n

Imax⁡Imin⁡=(I1+I2I1−I2)2{{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}Imin​Imax​​=(I1​​−I2​​I1​​+I2​​​)2

= (I1I2+1I1I2−1)2{\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}(I2​I1​​​−1I2​I1​​​+1​)2 = (n+1n−1)2{\left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)^2}(n​−1n​+1​)2

Imax⁡−Imin⁡Imax⁡+Imin⁡={{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}} = Imax​+Imin​Imax​−Imin​​= Imax⁡Imin⁡−1Imax⁡Imin⁡+1{{{{{I_{\max }}} \over {{I_{\min }}}} - 1} \over {{{{I_{\max }}} \over {{I_{\min }}}} + 1}}Imin​Imax​​+1Imin​Imax​​−1​

= (n+1n−1)2−1(n+1n−1)2+1{{{{\left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)}^2} - 1} \over {{{\left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)}^2} + 1}}(n​−1n​+1​)2+1(n​−1n​+1​)2−1​

= (n+1)2−(n−1)2(n+1)2+(n−1)2{{{{\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}}}(n​+1)2+(n​−1)2(n​+1)2−(n​−1)2​

= 4n2(n+1){{4\sqrt n } \over {2\left( {n + 1} \right)}}2(n+1)4n​​ = 2nn+1{{2\sqrt n } \over {n + 1}}n+12n​​

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