To an eye placed at a depth below calm surface of refractive index ų in a lake the surface appears like a reflecting plane mirror with a circular hole of radius R through which objects situated outside water can be seen.Then R is equal to
A).h
B)h/ų
C)ųh
D)h/ųsquare-1
Answers
Answer:
Explanation:
Surface of refractive index = ų (Given)
Radius of the circular hole = R (Given)
let the radius of the circular area = x
Therefore,
x/h= tan c
x/h= sin c/√(1- sin ²c) = 1/μ/√(1-1/μ²)
since sin c= 1/μ
Now,
x/h = 1/√μ²-1
x = h/√μ²-1
Hence, the light escapes are through the water directly above the point source
Thus, the angle subtended by a radius of the area on the source will be -
c= sin⁻¹(1/μ)
Answer:
Explanation:
Surface of refractive index = ų (Given)
Radius of the circular hole = R (Given)
let the radius of the circular area = x
Therefore,
x/h= tan c
x/h= sin c/√(1- sin ²c) = 1/μ/√(1-1/μ²)
since sin c= 1/μ
Now,
x/h = 1/√μ²-1
x = h/√μ²-1
Hence, the light escapes are through the water directly above the point source
Thus, the angle subtended by a radius of the area on the source will be -
c= sin⁻¹(1/μ)
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