Question

The critical angle for total internal reflection at a liquid-air interface is 42.0 ∘. if a ray of light traveling in the liquid has an angle of incidence at the interface of 30.5 ∘, what angle does the refracted ray in the air make with the normal? if a ray of light traveling in air has an angle of incidence at the interface of 30.5 ∘, what angle does the refracted ray in the liquid make with the normal?

Answer 1

The angle of refraction in the air can be found using Snell's law and the critical angle. For an angle of incidence of 30.5° in the liquid, the angle of refraction in the air can also be found with Snell's law.

Step-by-step explanation:

When light travels from a denser medium (liquid) to a less dense medium (air), total internal reflection occurs if the angle of incidence is greater than the critical angle. In this case, the critical angle is 42.0°.

If a ray of light traveling in the liquid has an angle of incidence of 30.5° at the interface, it is less than the critical angle. Therefore, it will be refracted (bent) away from the normal. To find the angle of refraction, we need to use Snell's law:

n1sinθ1 = n2sinθ2

Where n1 and n2 are the refractive indices of the liquid and air respectively.

In this case, we know the angle of incidence (θ1) is 30.5°. Assuming the refractive index of air is 1, we can find the refractive index of the liquid:

1.00 * sin(30.5°) = n2 * sin(42.0°)

Solving this equation, we can find the refractive index of the liquid. Once we have that, we can find the angle of refraction (θ2) using Snell's law.

For the second part of the question, if a ray of light traveling in air has an angle of incidence of 30.5°, the same process can be followed but this time we are finding the angle of refraction in the liquid.

Answer 2

critical angle = the 42 degrees means when the ray of light drifting in liquid is occurrence at 42 deg with normal, the refracted ray just scratches the surface of liquid.
sin i / sin r = u (refractive index of air with respect to liquid)
so sin 42 / sin 90 = u since sin 90 =1 therefore, u of liquid w.r.t. air = sin 42 deg.
Case 1 since /_ i = 30.5 deg sin 30.5/ sin r = u (which is sin 42 deg) so sin r = sin 31 / u angle r = sin inverse of (sin 30.5 /sin 42) angle r = 49.33



Case 2 the ray travels from air to liquid
So the refractive index of liquid wrt air = 1/u
angle i = 30.5 angle r =? sin 30.5 / sin r = 1/u = 1/sin 42 so, sin r = sin 30.5 x sin 40 angle r = arc sin (sin 30.5 x sin 40)angle r = 19.04 degrees