Question

Light enters from air into a given medium at an angle of \( 45^\circ \) with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of \( 15^\circ \) from its original direction. The refractive index of the medium is

• A. \( 1.732 \)
• B. \( 1.333 \)
• C. \( 1.414 \)
• D. \( 2.732 \)

Hint:

Understand the definitions of the angle of incidence, angle of refraction, and angle of deviation. Use the relationship between these angles. Apply Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media. Remember the values of trigonometric functions for common angles like \( 30^\circ \) and \( 45^\circ \).

Answer 1

The Correct Option is C

The angle of incidence \( i \) is the angle between the incident ray and the normal to the surface.
Given \( i = 45^\circ \).
The angle of deviation \( \delta \) is the angle between the direction of the incident ray and the direction of the refracted ray.
Given \( \delta = 15^\circ \).
The angle of refraction \( r \) is the angle between the refracted ray and the normal to the surface.
Since the light deviates towards the normal when entering a denser medium (as implied by refraction), we have \( r = i - \delta \).
\( r = 45^\circ - 15^\circ = 30^\circ \).
The refractive index \( n \) of the medium is given by Snell's law: $$ \frac{\sin i}{\sin r} = n $$ Here, the light is going from air (refractive index approximately 1) to the medium with refractive index \( n \).
$$ \frac{\sin 45^\circ}{\sin 30^\circ} = n $$ We know that \( \sin 45^\circ = \frac{1}{\sqrt{2}} \) and \( \sin 30^\circ = \frac{1}{2} \).
$$ n = \frac{1/\sqrt{2}}{1/2} = \frac{1}{\sqrt{2}} \times \frac{2}{1} = \frac{2}{\sqrt{2}} = \sqrt{2} $$ The value of \( \sqrt{2} \) is approximately \( 1.
414 \).
Therefore, the refractive index of the medium is \( 1.
414 \).