Question

an opaque cylindrical tank with an open top has a diameter of3.00 m and is completely filled with water. when theafternoon Sun reaches an angle of 28.0 degrees above the horizon,sunlight ceases to illuminate any part of the bottom of thetank. how deep is the tank?

Answer 1

When light passes from one medium to another it bends due torefraction.

So your question can be picturised as follows,

when the afternoon Sun reaches an angle of 28.0 degrees abovethe horizon, sunlight ceases to illuminate any part of the bottomof the tank , the bottom will remain dark

Here i is the angle of incidence and r is the angle ofrefraction.

i =90-28 =62 degrees.

Now we can apply snells law,

n₁sin i = n₂sin r . {n₁=1isthe refractive index of air and n₂=1.33 is that ofwater

sin r =sin 62/1.33 =0.66.   or   r=sin^-10.66 ˜ 41.3 degrees.

Thus we have obtained the angle of refraction.

Now we can easily obtain the depth of the tank usingtrignometry,

From figure,   tan r =3/d, where d is thedepth of the tank

or d =3/tan r=3.4m

Answer 2

Angle of incidence(between ray and normal) = 90-28 =62^o

Let Ref index of water, n = 1.33

Sin i/Sin r = n 1.33

Sin r = sin 62/1.33 = 0.8829/1.33 = 0.6639

Angle of refraction, r = 41.6

Tan r = D/d

Depth of the tank,

d = D/tan r = 3.0 /tan 41.6

                 = 3.0 /0.8877 = 3.38m