Problem 7 - Arihant AIEEE Physics

Question

A ray of light falls on a transparent glass slab of refractive index 1.62. What is the angle of incidence, if the reflected ray and refracted ray are mutually perpendicular? (a) \(\tan ^{-1}(1.62)\) (b) \(\tan ^{-1}\left(\frac{1}{1.62}\right)\) (c) \(\frac{1}{\tan ^{-1}(1.62)}\) (d) None of these

Step-by-Step Solution

Answer

The angle of incidence is \(\tan^{-1}(1.62)\); option (a) is correct.

Step 1: Analyze the Geometry

Based on the given problem, when a ray of light hits a transparent surface, it splits into a reflected ray and a refracted ray. These two rays being mutually perpendicular means the angle between them is 90 degrees.

Step 2: Apply the Refractive Index Information

The problem provides a refractive index of 1.62 for the glass slab. The formula for refractive index, when the incident and refracted angles are complementary (i.e., they add up to 90°), is \(\mu = \tan(i)\), where \(\mu\) is the refractive index and \(i\) is the angle of incidence.

Step 3: Solve Using Trigonometric Relationships

We have \(\mu = \tan(i)\), and substituting the given refractive index, the equation becomes \(\tan(i) = 1.62\). Therefore, \(i = \tan^{-1}(1.62)\).

Step 4: Review Options and Identify the Correct Answer

Compare the calculated solution \(i = \tan^{-1}(1.62)\) with the given options. Option (a) matches \(\tan^{-1}(1.62)\), making it the correct answer.

Key Concepts

Refractive Index

The refractive index is a measure of how much a material can bend light. When light travels from one medium to another, its speed changes. This bending of light is quantified using the refractive index, represented as \( \mu \). It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium.

• Mathematically, it is given by \( \mu = \frac{c}{v} \), where \( c \) is the speed of light in a vacuum and \( v \) is the speed of light in the medium.
• A higher refractive index indicates that light travels slower in the medium, resulting in bending towards the normal.

In the context of the exercise, the refractive index of the glass slab is 1.62. This value helps us determine how much the light ray will refract (bend) when entering the glass. By understanding the refractive index, we can predict changes in light paths and solve related geometric problems in optics.

Angle of Incidence

The angle of incidence is the angle between the incoming light ray and the normal line at the surface where the light ray strikes. It plays a crucial role in determining how light rays behave when hitting different surfaces.

• The angle of incidence is often denoted as \(i\).
• When light hits a surface, it can either be reflected or refracted, depending on the properties of the surface and the angle.

In the given exercise, the angle of incidence determines how the light ray is split into a reflected and refracted ray. Since the problem states that these two rays are perpendicular, it allows us to use the relationship \( \mu = \tan(i) \) to precisely find the value of \(i\). Here, we are given a refractive index of 1.62, leading to the conclusion that \(i = \tan^{-1}(1.62)\). This establishes the angle of incidence needed to create 90-degree mutual angles between the reflected and refracted rays.

Trigonometric Relationships

Trigonometric relationships are vital in solving problems involving angles and distances in optics. In particular, the tangent function is frequently used because it connects the angle with the perpendicular sides of a right-angled triangle.For the exercise, the relationship \( \mu = \tan(i) \) connects the refractive index with the angle of incidence:

• \( \tan(i) = \frac{\text{opposite}}{\text{adjacent}} \)
• The inverse tangent, or \( \tan^{-1} \), helps to find the angle when the refractive index is known.

Given \( \mu = 1.62 \), we solve for the angle of incidence as \(i = \tan^{-1}(1.62)\). This function allows us to translate from a numerical refractive index to an actual angle measurement. By employing these trigonometric relationships, particularly the tangent and its inverse, we can solve complex problems involving light behavior and angle calculations.