Problem 116 - Master Resource Book in JEE Main Physics

Question

A ray \(O P\) of monochromatic light is incident on the face \(A B\) of prism \(A B C D\) near vertex. \(B\) at an incident angle of \(60^{\circ}\) (see figure). If the refractive index of the material of the prism is \(\sqrt{3}\), which of the following is (are) correet? [IIT JEF 2010] (a) The ray gets totally internally reflected at face \(\mathrm{CD}\) (b) The ray comes out through face \(A D\) (c) The angle between the incident ray and the emergent ray is \(90^{\circ}\) (d) The angle between the incident ray and the emergent ray is \(120^{\circ}\)

Step-by-Step Solution

Answer

Correct options: (a) and (d).

Step 1: Analyze Incident Ray

The incident ray enters the prism at an angle of \(60^{\circ}\) with respect to the normal at face \(AB\). Since the prism material's refractive index is \(\sqrt{3}\), use Snell's Law to find the angle of refraction.

Step 2: Apply Snell's Law

Snell's Law is given by \(n_1 \sin \theta_1 = n_2 \sin \theta_2\). Here, \(n_1 = 1\), \(\theta_1 = 60^{\circ}\), and \(n_2 = \sqrt{3}\). Thus, \(1 \cdot \sin 60^{\circ} = \sqrt{3} \cdot \sin \theta_2\). Solve for \(\theta_2\): \[\sin \theta_2 = \frac{\sin 60^{\circ}}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}.\] Therefore, \(\theta_2 = 30^{\circ}.\)

Step 3: Determine Path Inside Prism

When the light ray inside the prism reaches face \(CD\), determine if total internal reflection occurs. The critical angle \(\theta_c\) given \(n_1' = \sqrt{3}\) (inside) to \(n_2' = 1\) (outside) is calculated using: \[\sin \theta_c = \frac{1}{\sqrt{3}}.\] So, \(\theta_c \approx 35.26^{\circ}.\) The angle of incidence at face \(CD\) will be greater than \(\theta_c\), verifying total internal reflection.

Step 4: Total Internal Reflection

Since the angle of incidence at \(CD\) exceeds the critical angle, the ray undergoes total internal reflection and travels towards face \(AD\).

Step 5: Emerging Ray Analysis

After total internal reflection, calculate the angle of emergence at face \(AD\). Since the internal angle is the same due to the symmetric prism and the total internal reflection, apply Snell's Law reversely to find it emerges into the air at \(60^{\circ}.\)

Step 6: Calculate Angle Between Incident and Emergent Ray

The emergent ray exits at \(60^{\circ}\) along face \(AD\), total deviation inside the prism is \(60^{\circ} + 60^{\circ} = 120^{\circ}.\)

Step 7: Conclusion Based on Calculations

Combine the results: choice (a) is correct as total internal reflection occurs, and choice (d) is correct since the incident and emergent rays have a \(120^{\circ}\) deviation.

Key Concepts

Snell's Law

Snell's Law is a fundamental principle in optics that describes how light travels through different media. It relates the angle of incidence and refraction for a light ray crossing the boundary between two different isotropic media, such as air and glass. This law is usually written as:

• \[n_1 \sin \theta_1 = n_2 \sin \theta_2\]

Here, \( n_1 \) and \( n_2 \) are the refractive indices of the first and second media, while \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.
Snell's Law helps us understand how light bends when it enters a new medium. For instance, in our problem, light travels from air (with \( n_1 = 1 \)) into a prism material (with \( n_2 = \sqrt{3} \)).
By applying Snell's Law, we calculated the angle of refraction \( \theta_2 \) to be \(30^{\circ}\). This calculation shows how the light ray changes direction when entering the prism.
Understanding how angles and refractive indices influence light's path is crucial for numerous applications, including lenses, prisms, and fiber optics.

Angle of Deviation

The angle of deviation is an important concept in understanding how light interacts with devices like prisms. It refers to the angle through which a light ray deviates, or bends, as it passes through a prism.
In simple terms, the angle of deviation is the difference between the direction of the original incident ray and the final emergent ray.

• For example, if a ray of light enters one face of a prism and finally exits through another face, it will experience a change in direction.
• This total change is the angle of deviation.

In our problem, the deviation is critical for comprehending how light navigates a prism.
Initially, the incident ray enters at an angle of \(60^{\circ}\), and due to total internal reflection, the deviation results in the emergent ray exiting at \(60^{\circ}\).
Adding the paths inside the prism gives us a complete deviation of \(120^{\circ}\). This information is vital for optical system design and analysis because it tells us how much the light's path would be shifted by the prism.

Critical Angle

The critical angle is a key concept in the phenomenon of total internal reflection, where a light ray gets completely reflected inside a medium instead of refracting through the boundary. This occurs when light moves from a medium with a higher refractive index to one with a lower refractive index.
The critical angle \( \theta_c \) can be calculated by:

• \(\sin \theta_c = \frac{n_2}{n_1}\)

Here, \(n_1\) is the refractive index of the denser medium, and \(n_2\) is that of the less dense medium.
In the exercise, the critical angle is calculated for light traveling from the prism material \((n_1 = \sqrt{3} )\) back to air \((n_2 = 1 )\).
The critical angle calculated is approximately \(35.26^{\circ}\).

• Any incident angle greater than this results in total internal reflection.
• This concept explains why the light stays inside the prism at face CD.

The understanding of critical angles is crucial in designing optical devices like binoculars and optical fibers, where light needs to stay within a medium.