Question

Let's apply Snell's law to the refraction of light across a water–air interface. Suppose you kneel beside the fishpond in your backyard and look at one of the fish. You see it by sunlight that reflects off the fish and refracts at the water–air interface. If the light from the fish to your eye strikes the water–air interface at an angle of 60.0∘∘ to the interface, what is the angle of refraction of the ray in the air?

Answer 1

Answer:

The angle of refraction is 41.68°.

Explanation:

The refractive index for water is $n_2 = 1.333$, and for air $n_1 = 1.00$: the angle of light with the normal is $90^o-60^o = 30^o$; therefore Snell's law gives

$n_1sin(\theta_1)= n_2sin(\theta_2)$

$1.00*sin(\theta_1) = 1.33 sin(30^o)$

$sin (\theta_1) = \dfrac{1.33sin(30^o)}{1.00}$

$sin (\theta_1) = 0.665$

$\theta _1 = sin^{-1}(0.665)$

$\boxed{\theta_1 = 41.68^o}$