A point source is placed at a depth h below the surface of water (refractive index = μ). (a) Show that light escapes through a circular area on the water surface with its centre directly above the point source. (b) Find the angle subtended by a radius of the area on the source.
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Given :
let x be the radius of circular area
x/h= tan c
x/h= sin c/√(1- sin ²c) = 1/μ/√(1-1/μ²)
since sin c= 1/μ
x/h= 1/√μ²-1
x= h/√μ²-1
so light escapes through are on the water directly above the point sorce
b) angle subtended by a radius of the area on the source
c= sin⁻¹(1/μ)
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Answer is given below....
let x be the radius of circular area
x/h= tan c
x/h= sin c/√(1- sin ²c) = 1/μ/√(1-1/μ²)
since sin c= 1/μ
x/h= 1/√μ²-1
x= h/√μ²-1
so light escapes through are on the water directly above the point sorce
b) angle subtended by a radius of the area on the source
c= sin⁻¹(1/μ)
let x be the radius of circular area
x/h= tan c
x/h= sin c/√(1- sin ²c) = 1/μ/√(1-1/μ²)
since sin c= 1/μ
x/h= 1/√μ²-1
x= h/√μ²-1
so light escapes through are on the water directly above the point sorce
b) angle subtended by a radius of the area on the source
c= sin⁻¹(1/μ)
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