A fisherman sees rays of light apparently diverging from a p

A fisherman sees rays of light apparently diverging from a point P\' on a fish (see figure below). The fish appears to be aty = 0.90 m below the surface. What is its actual depth?

(Take = 49.4°.)

Solution

tan (theta) = d /y

d = 0.9 tan49.4 = 1.049 m

and incident angle = 49.4 deg

Using snell\'s law to find rfraction angle,

ni sini = nr sinr (refractive index of water, nr = 1.33 )

1 * sin 49.4 =1.33 sinr

0.7592/1.33 = sinr

=> r = 34.75 deg

and tan 34.75 = d / y\"

y\" = 1.049 /0.693 = 1.5137 m