Question : A luminous object is located at the bottom of a big pool of liquid of refractive index µ and depth h. The object S emits rays upward in all directions so that a circle of light is formed at the surface of the liquid by the rays which are refracted into the air. What happened to the rays beyond the circle? Determine the radius and area of the circle.
Doubt by Sonali
Solution :
Rays, which are beyond the circle, will be refracted out of water.
In Rt. ΔAOS
OA = r
OS = h
OS = h
AS²=OA²+OS² (By Pythagoras Theorem)
AS²=r²+h²
AS=√[r²+h²]
Again, In Rt. ΔAOS
Sinic=P/H
Sinic=OA/AS
Sinic=r/(√r²+h²) — (1)
Also,
µ = 1/Sinic
µ = 1/[r/(√r²+h²)] [∵ Using equation (1)]
µ = (√r²+h²)/r
µr=√r²+h²
Squaring both sides
µ²r² = r²+h²
µ²r²-r²=h²
r²(µ²-1)=h²
r²=h²/(µ²-1)
µ = 1/Sinic
µ = 1/[r/(√r²+h²)] [∵ Using equation (1)]
µ = (√r²+h²)/r
µr=√r²+h²
Squaring both sides
µ²r² = r²+h²
µ²r²-r²=h²
r²(µ²-1)=h²
r²=h²/(µ²-1)
r²=h/√(µ²-1)
Area of the circle =πr² =π[h/√(µ²-1)]² =π[h²/(µ²-1)]
=πh²/(µ²-1)
