Question

A transparent object with an isosceles right triangular cross-section has an index of refraction n2 = 1.2, as shown in the diagram below. A transparent object with an isosceles right trian

A light beam in air is incident on this object, making an angle ?in = 75? with respect to the x-axis, as shown. At what angle (with respect to the x-axis), ?out, does the observer see the light beam exit the object?

Answer 1

The observer sees the light beam exit the object at an angle of approximately 63.53 degrees.

Step-by-step explanation:

The question asks for the angle at which the observer sees a light beam exit a transparent object with an isosceles right triangular cross-section. To determine this angle, we can use Snell's law, which relates the angles and indices of refraction of light as it passes from one medium to another. Snell's law is given by: n1 sin(θ1) = n2 sin(θ2) where n1 and n2 are the indices of refraction of the initial and final mediums, respectively, and θ1 and θ2 are the angles of incidence and refraction, respectively.

In this case, the light beam enters the object from air, where the index of refraction is 1.00, and exits into the object, where the index of refraction is 1.2. The angle of incidence is given as 75 degrees. Let's denote the angle of refraction as ?out. Using Snell's law, we can write: 1.00 * sin(75) = 1.2 * sin(?out)

Solving this equation for ?out, we find that the observer sees the light beam exit the object at an angle of approximately 63.53 degrees.