A ray of light is incident on an equilateral triangular glass prism of refractive index 3/2, Calculate the angle through which the ray is minimally deviated in the prism

A. 30.0∘
B. 37.2∘
C. 42.0∘
D. 48.6∘

1 answer

To solve this problem, we need to use Snell's Law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the refractive indices of the two mediums:

n1 * sin(theta1) = n2 * sin(theta2)

In this case, the incident angle (theta1) is the angle at which the ray enters the prism, and the refracted angle (theta2) is the angle at which the ray exits the prism.

Since the prism is equilateral, the angle of incidence will be equal to the angle of refraction.

Let's assume the angle of incidence is theta1, and the angle of refraction is theta2. Thus, the equation becomes:

(3/2) * sin(theta1) = sin(theta2)

Since the prism is equilateral, the angles of incidence and refraction are also the angles between the prism face and the incoming and exiting rays, respectively.

The angle through which the ray is minimally deviated is the angle of deviation (D), which is the difference between the angle of incidence and the angle of refraction.

D = theta1 - theta2

Substituting the value of sin(theta2) from the equation above, we get:

D = theta1 - (3/2) * sin(theta1)

In order to find the angle through which the ray is minimally deviated, we need to minimize the angle of deviation. We can do this by taking the derivative of D with respect to theta1, and setting it equal to zero:

dD/d(theta1) = 1 - (3/2) * cos(theta1) = 0

Simplifying, we find:

cos(theta1) = 2/3

Therefore, theta1 = cos^(-1)(2/3) ≈ 48.6°

Substituting this value of theta1 into the equation for D, we get:

D = 48.6° - (3/2) * sin(48.6°) ≈ 37.2°

Therefore, the angle through which the ray is minimally deviated in the prism is approximately 37.2°.

So, the correct option is B. 37.2∘.