Please help me in this. How to do it?

Question

A girl sees through a circular slab #(mu_g=1.5)# of thickness 20 mm and diameter 60 cm, to the bottom of a swimming pool. The bottom surface of the slab is in contact with the water surface #(mu_w=1.33)#. The depth of swimming pool is 6 m. What is the area of bottom of swimming pool that can be seen through the slab?

Thank you!

Sebastian Acosta · Accepted Answer

<mathjax>#160m^2# <img src="https://tutor.hix.ai 
For maximum area of swimming pool to be seen through glass slab, let us consider a ray of light (shown in red) starting from the bottom and hitting the glass slab at its edge. As it travels from rarer to denser medium the ray bends towards the normal. 
Applying Snell's law at this interface A 
<mathjax>#n_1sini=n_2sinr#</mathjax> 
<mathjax>#1.33sini=1.5sinr#</mathjax> .......(1)
After the ray travels through the glass slab and hits interface B, it is refracted again. 
From Geometry we see that now the angle of incidence is <mathjax>#=angler#</mathjax>. 
As the ray travels from denser to rarer medium, it is bent away from the normal. 
For maximum view of the bottom of the pool, angle of refraction <mathjax>#=90^@#</mathjax> 
Applying Snell's law at this interface we get 
<mathjax>#1.5sinr=1sin90^@#</mathjax> 
<mathjax>#=>sinr=1/1.5#</mathjax> .......(2) 
From (1) and (2) we get 
<mathjax>#1.33sini=1.5xx1/1.5=1#</mathjax> 
<mathjax>#sini=1/1.33#</mathjax> ......(3)
Making use of trigonometry to calculate <mathjax>#x#</mathjax>. 
In the relevant triangle 
<mathjax>#x/6=tani#</mathjax> 
<mathjax>#=>x=6xxsini/cosi=6xx(sini)/sqrt(1-sin^2i)#</mathjax> 
Using (3) we get 
<mathjax>#x=6xx(1/1.33)/sqrt(1-(1/1.33)^2)#</mathjax> 
<mathjax>#=>x=6.84m#
Radius of visible circle<mathjax>#=0.3+x=7.14m#
Visible area of the bottom <mathjax>#=pi(7.14)^2=160m^2#

Carson Alexander · Answer

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