A cylindrical cistern, constructed below ground level, is 2.9 m in diameter and 2.0 m deep and is filled to the brim with a liquid whose index of refraction is 1.4. A small object rests on the bottom of the cistern at its center. How far from the edge of the cistern can a girl whose eyes are 1.2 m from the ground stand and still see the object

15.1 m

Explanation:

We first calculate the apparent depth from

refractive index, n = real depth/apparent depth

apparent depth, a  = real depth/refractive index

real depth = 2.0 m, refractive index = 1.4

apparent depth, a = 2.0/1.4 = 1.43 m

Since the cylindrical cistern has a diameter of 2.9 m, its radius is 2.9/2 = 1.45 m

The angle of refraction, r is thus gotten from the ratio

tan r = radius/apparent depth = 1.45/1.43 = 1.014

r = tan⁻¹(1.014) = 45.4°

The angle of incidence, i is gotten from n = sin i/sin r

sin i = nsin r = 1.4sin45.4° = 1.4 × 0.7120 = 0.9968

i = sin⁻¹(0.9968) = 85.44°

Since the girl's eyes are 1.2 m from the ground, the distance ,h from the edge of the cistern she must stand is gotten from

tan i = h/1.2

h = 1.2tan i = 1.2tan85.44° = 1.2 × 12.54 = 15.05 m

h = 15.05 m ≅ 15.1 m

So, she must stand 15.1 m away from the edge of the cistern to still see the object.

it is a negative acceleration