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There is a small source of light at some depth below the surface of water (refractive index =\frac{4}{3} ) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly) : [ Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2\pi rh ]
Option: 1 34\; ^{o}/_{o}      
Option: 2 17\; ^{o}/_{o}      
Option: 3 50\; ^{o}/_{o}      
Option: 4 21\; ^{o}/_{o}      
 

Answers (1)

best_answer

 

 

  

\sin \beta=\frac{3}{4},\cos \beta =\frac{\sqrt{7}}{4}

Solid\; angle\; d\Omega =2\pi R^{2}(1-\cos \beta )

Percentage of light =\frac{2\pi R^{2}(1-\cos \beta )}{4\pi R^{2}}\times 100=\frac{1-\cos \beta }{2}\times 100=\left ( \frac{4-\sqrt{7}}{8} \right )\times 100\approx 17\: ^{o}/_{o}

Hence the correct option is (2).

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avinash.dongre

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