Wave Optics

JEE Main level, single correct, medium

72 of 191

Question 72 of 191

Two coherent sources of light interfere. The intensity ratio of two sources is 1:41:4. For this interference pattern if the value of Imax⁡+Imin⁡Imax⁡−Imin⁡\dfrac{{I}_{\max }+{I}_{\min }}{{I}_{\max }-{I}_{\min }} is equal to 2α+1β+3\dfrac{2\alpha +1}{\beta +3}, then αβ\dfrac{\alpha }{\beta } will be

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Answer: (B) 22

Given here, I1I2=14\dfrac{{I}_{1}}{{I}_{2}}=\dfrac{1}{4} or I2=4I1{I}_{2}=4{I}_{1}
Maximum intensity is Imax⁡=(I1+I2)2=I1+4I1+2I14I1=9I1{I}_{\max }={\left(\sqrt{{I}_{1}}+\sqrt{{I}_{2}}\right)}^{2}={I}_{1}+4{I}_{1}+2\sqrt{{I}_{1}4{I}_{1}}=9{I}_{1}
Minimum intensity is Imin⁡=(I1−I2)2=I1+4I1−2I14I1=I1{I}_{\min }={\left(\sqrt{{I}_{1}}-\sqrt{{I}_{2}}\right)}^{2}={I}_{1}+4{I}_{1}-2\sqrt{{I}_{1}4{I}_{1}}={I}_{1}
The ratio of Imax⁡+Imin⁡Imax⁡−Imin⁡\dfrac{{I}_{\max }+{I}_{\min }}{{I}_{\max }-{I}_{\min }} is
9I1+I19I1−I1=108=54=2α+1β+3\dfrac{9{I}_{1}+{I}_{1}}{9{I}_{1}-{I}_{1}}=\dfrac{10}{8}=\dfrac{5}{4}=\dfrac{2\alpha +1}{\beta +3}
α=2β=1\alpha =2\beta =1
∴αβ=21=2∴\dfrac{\alpha }{\beta }=\dfrac{2}{1}=2

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