Problem 10 - Master Resource Book in JEE Main Physics

Question

The \(x-y\) plane is the boundary between two transparent media. A medium I has a refractive index \(\mu_{1}=\sqrt{2}\) and medium II has a refractive index \(\mu_{2}=\sqrt{3}\). A ray of light in medium I, given by vector, \(\mathbf{A}=\sqrt{3} \hat{\mathbf{i}}-\hat{\mathbf{k}}\) is incident on the plane of separation. The unit vector in the direction of the refracted ray in medium II is (a) \(\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{k}})\) (b) \(\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\) (c) \(\frac{1}{\sqrt{2}}(\hat{\mathrm{k}}-\hat{\mathrm{i}})\) (d) \(\frac{1}{\sqrt{2}}(\hat{i}-\hat{k})\)

Step-by-Step Solution

Answer

The unit vector direction is (d) \( \frac{1}{\sqrt{2}}(\hat{i} - \hat{k}) \).

Step 1: Identify Incident Vector Components

The given incident ray vector is \( \mathbf{A} = \sqrt{3} \hat{\mathbf{i}} - \hat{\mathbf{k}} \). This vector indicates that in medium I, the ray travels with a component \( \sqrt{3} \) along the \( \hat{\mathbf{i}} \) direction and \( -1 \) along the \( \hat{\mathbf{k}} \) direction. Since the ray is incident in the \(x-y\) plane (which is perpendicular to the \(z\)-axis), the components reveal the angle with respect to the normal.

Step 2: Calculate Incident Angle with Normal

The incident vector \( \mathbf{A} \) is in the \(xz\)-plane, and the incident angle \( \theta_1 \) with the normal (\( \hat{\mathbf{k}} \)) can be calculated by resolving the vector. We see \( \tan \theta_1 = \frac{\sqrt{3}}{1} = \sqrt{3} \), which suggests \( \theta_1 = 60^\circ \).

Step 3: Apply Snell's Law

According to Snell's Law, \( \mu_1 \sin \theta_1 = \mu_2 \sin \theta_2 \). Given \( \mu_1 = \sqrt{2} \), \( \theta_1 = 60^\circ \), and \( \mu_2 = \sqrt{3} \), therefore \( \sqrt{2} \cdot \sin 60^\circ = \sqrt{3} \cdot \sin \theta_2 \). Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), substituting gives \( \sqrt{2} \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin \theta_2 \). Simplifying \( \sin \theta_2 = \frac{\sqrt{2} \cdot \sqrt{3}}{\sqrt{3} \cdot 2} = \frac{\sqrt{2}}{2} = \sin 45^\circ \) indicates that \( \theta_2 = 45^\circ \).

Step 4: Determine Refracted Ray Direction

As \( \theta_2 = 45^\circ \), the refracted ray should follow the laws of refraction and stay within the same plane (the \(xz\)-plane since \(y\)-component was absent in the incident ray). The ray refracts away from the normal. Thus its direction is a linear combination where the components have equal absolute values, reflecting \( \tan 45^\circ = 1 \).

Step 5: Match with Given Options

The refracted direction based on trigonometric values is given by \( \frac{1}{\sqrt{2}}( \hat{i} - \hat{k} ) \). This corresponds to option (d). Thus, the refraction maintains the \( \hat{i} \) and \( \hat{k} \) components equally balanced.

Key Concepts

Refractive Index

Refractive index is a fundamental concept in the study of optics. It measures how much a ray of light bends, or refracts, when it enters a medium from another. This concept is crucial for understanding how light behaves as it passes through different materials. The refractive index, denoted as \( \mu \), is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. Mathematically, it is written as:

• \( \mu = \frac{c}{v} \)

Here, \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium.
The refractive index of a medium affects how much light bends when crossing the boundary between two media. A higher refractive index indicates that light travels slower in that medium, resulting in a greater bending effect. In the exercise, medium I has a refractive index of \( \sqrt{2} \) and medium II has \( \sqrt{3} \).
This difference in refractive indices plays a critical role as Snell's Law relates the angle of incidence to the angle of refraction based on these indices.

Angle of Refraction

The angle of refraction is the angle formed between the refracted light ray and the normal (an imaginary line perpendicular to the interface of the two media). When light transitions between different media, its speed changes, resulting in a change of direction, or refraction.
According to Snell's Law, the relationship between the angles and the refractive indices of the two media is given by:

• \( \mu_1 \sin \theta_1 = \mu_2 \sin \theta_2 \)

Here, \( \theta_1 \) is the incident angle, and \( \theta_2 \) is the angle of refraction. In the provided exercise, when light travels from medium I to medium II, it refracts according to their refractive indices. The calculation shows that with \( \theta_1 = 60^\circ \) and the given refractive indices, the angle of refraction \( \theta_2 \) is calculated to be \(45^\circ\). This means the light bends away from the normal because medium II has a higher refractive index.

Incident Angle

The incident angle is the angle between the incoming light ray, or incident ray, and the normal. Understanding this angle is important because it determines how much the light will bend when entering a different medium.
In the exercise, the incident angle \( \theta_1 \) can be found from the given direction vector. The vector \( \mathbf{A} = \sqrt{3} \hat{\mathbf{i}} - \hat{\mathbf{k}} \) shows the components of the light in medium I.
By resolving this vector, the tan of the angle \( \theta_1 \) can be computed as:

• \( \tan \theta_1 = \frac{\sqrt{3}}{1} = \sqrt{3} \)

From this tan value, it indicates that \( \theta_1 = 60^\circ \). Knowing \( \theta_1 \) is crucial for applying Snell's Law and finding the angle of refraction. This tells us how the light will behave as it passes from one medium to another.