Question

A point source of light is placed at the bottom of the water lake. If the area of the illuminated circle on the surface is equal to 3 times the square of the depth of the lake. The refractive index of water is :
A. $\sqrt {\dfrac{\pi }{3} + 1} $
B. $\sqrt {\dfrac{\pi }{3}} + 1$
C. $\dfrac{\pi }{3} + 1$
D. $\dfrac{\pi }{4} + 1$

Answer 1

Hint: Point source is kept at the bottom of the lake and hence it emits rays in conical direction. Now these rays emit out so a person can view point source in water. Apparent distance of that point in the lake will be less than actual distance and that seems to be nearer to the observer. If these rays hit at less than critical angle then this happens. If it hits at a critical angle then circle shape will be formed on the surface and that’s the area mentioned in the question. Now by applying snell’s law we find out the answer.

Formula used: $\mu \sin (\theta ) = 1\sin ({90^0})$

Complete step by step answer:
Let refractive index of water be $\mu $
Refractive index of air is 1
And ϴ is the critical angle for water air media
According to the snell’s law
$\mu \sin (\theta ) = 1\sin ({90^0})$

So $\mu = \dfrac{1}{{\sin (\theta )}}$ …eq1
From diagram
$\tan (\theta ) = \dfrac{r}{h}$
Where ‘r’ is the radius of the circle formed and ‘h’ is the depth of the lake
Area of the circle formed is $\pi {r^2}$
And according to question $\pi {r^2} = 3{h^2}$
We can apply trigonometric ratios in order to find out the tan and sin values to find out the refraction index of the water
according using equation 1 we will find out $\dfrac{1}{{\sin (\theta )}}$ to find out refractive index as both are equal
$\eqalign{
  & \Rightarrow \dfrac{r}{h} = \sqrt {\dfrac{3}{\pi }} \cr
  & \Rightarrow \tan (\theta ) = \sqrt {\dfrac{3}{\pi }} \cr
  & \Rightarrow \sin (\theta ) = \sqrt {\dfrac{3}{{3 + \pi }}} \cr
  & \Rightarrow \dfrac{1}{{\sin (\theta )}} = \sqrt {\dfrac{{3 + \pi }}{3}} = \sqrt {1 + \dfrac{\pi }{3}} \cr
  & \Rightarrow \mu = \dfrac{1}{{\sin (\theta )}} = \sqrt {1 + \dfrac{\pi }{3}} \cr} $

So, the correct answer is “Option A”.

Note: Here if the rays come out of the water then observer can see point source only and if rays completely reflects into water it self observer can’t see anything. Hence to form a circle on the surface rays should exactly hit at a critical angle and this is to be understood not given in the problem.