Question

In searching the bottom of a pool at night, a security guard shines a narrow beam of light from his flashlight, above the water's surface, onto the waiter at a point 2.5 m from his foot at the edge of the pool. Where does the light hit the bottom of the pool measured from the wall beneath his foot if the pool is 2.1 m deep. A triangular prism made of crown glass, n = 1.52, with base angles of 30 degree is surrounded by air. If parallel rays are indecent normally on its base, what is the angle phi between the two emerging rays? What is phi if the prism is emerged in water, n = 1.47?

Answer 1

Given,

h = 1.3 m and d = 2.5 m ; D = 2.1 m

The angle of incidence of light will be given by:

tan $\theta$ = d/h = 2.5/1.3 = 1.923 => $\theta$ = tan -1 ( 1.923)

$\theta$ = 62.53 degrees.

As the light is travelling from the air into water, we can calculate the angle of refraction using Snell's law.

n1 sin $\theta$ 1 = n2 sin $\theta$ 2

1 x sin62.53 = 1.33 x sin $\theta$ 2

$\theta$ 2 = 41.84 degrees

Now the horizontal distance from the edge of the pool will be :

L = d + D tan $\theta$ 2 = 2.5 + 2.1 x tan41.84 = 4.38 m

Hence, L = 4.38 m.