Question

A fisherman sees rays of light apparently diverging from a point P' on a fish (see figure below). The fish appears to be at y = 1.1 m below the surface. What is its actual depth? (Take ? = 47.1°.) The answer is not 1.79 m....

Answer 1

When fish is at F it appears to him as if it is at F'

So, if PF is the actual depth, P'F' is the apparent depth which is given as 1.1 m

Angle of refraction FO1N1 = PFO1 = 47.1 degrees

Now applying snell's law at point O1,

Sin i1 /Sin r1 = n

Sin i1 / Sin 47.1 = 1.33

Sin i1 = 1.815593329

i1= 76.97760268 degrees

So, angle N'1O1R1 is 76.97760268 degrees and hence angle R1O1H = P'O1F' = 90 - 76.97760268 = 13.02239732 degrees

similarly using the second ray, we can find angle P'O2F1 as 8.136232768 degrees

Now in triangle P'F'O1 , tan  P'O1F' = P'F' /P'O1

P'O1 = P'F' / tan 13.02239732 = 1.1 / tan 13.02239732 = 4.756140345 m

similarly in triangle P'O2F' , P'O2 = P'F' / tan 8.136232768 = 1.1 / tan 8.136232768 =7.694119629 m

So, O1O2 = P'O2 - P'O1 = 7.694119629 - 4.756140345 = 2.937979284 m

Now in triangle PFO1,

tan PFO1 = PO1 / PF

PF = PO1 / tan PFO1

PO1 = d tan 47.1

similarly, from triangle PFO2,

PO2 = PF tan PFO2 = d tan 48.1

PO2 - PO1 = d tan 48.1 - d tan 47.1

O1O2 = d ( tan 48.1 - tan 47.1 )

2.937979284 = d ( 0.038390049)

d = 76.52970901 m

So, the actual depth PF is d = 76.53 m