Problem 106 - Master Resource Book in JEE Main Physics

Question

Let the \(x z\)-plane be the boundary between two transparent media. Medium 1 in \(z \geq 0\) has a refractive index of \(\sqrt{2}\) and medium 2 with \(z<0\) has a refractive index \(\sqrt{3} .\) A ray of light in medium 1 given by the vector \(\mathbf{A}=6 \sqrt{3} \hat{\mathbf{i}}+8 \sqrt{3} \hat{\mathbf{j}}-10 \hat{\mathbf{k}}\) is incident on the plane of separation. The angle of refraction in medium 2 is (a) \(45^{*}\) (b) \(60^{\circ}\) (c) 75 (d) \(30^{*}\)

Step-by-Step Solution

Answer

The angle of refraction in medium 2 is \( 60^\circ \).

Step 1: Find the Incident Angle Using Vector

To find the incident angle, we first calculate the direction cosine of vector \( \mathbf{A} = 6 \sqrt{3} \hat{\mathbf{i}} + 8 \sqrt{3} \hat{\mathbf{j}} - 10 \hat{\mathbf{k}} \). The incident angle \( \theta_i \) is the angle between \( \mathbf{A} \) and the positive \( z \)-axis \( \hat{\mathbf{k}} \). The cosine of \( \theta_i \) is computed by finding the dot product of \( \mathbf{A} \) and \( \hat{\mathbf{k}} \), and then dividing by their magnitudes:\[ \cos \theta_i = \frac{\mathbf{A} \cdot \hat{\mathbf{k}}}{||\mathbf{A}||} = \frac{-10}{||\mathbf{A}||} \]Calculate \( ||\mathbf{A}|| = \sqrt{(6\sqrt{3})^2 + (8\sqrt{3})^2 + (-10)^2} = \sqrt{108 + 192 + 100} = \sqrt{400} = 20 \).Then, \[ \cos \theta_i = \frac{-10}{20} = -0.5 \] which gives \( \theta_i = \cos^{-1}(-0.5) = 120^\circ \). However, since the angle needs to be between the ray and normal (z axis), adjust this to \( \theta_i' = 180^\circ - 120^\circ = 60^\circ \).

Step 2: Apply Snell's Law

Snell's law relates the angles and refractive indices of the two media. The law is given by:\[ n_1 \sin \theta_i = n_2 \sin \theta_t \]Here, \( n_1 = \sqrt{2} \), \( \theta_i = 60^\circ \), \( n_2 = \sqrt{3} \). Plug in these values to find \( \theta_t \):\[ \sqrt{2} \sin(60^\circ) = \sqrt{3} \sin \theta_t \]Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), the equation becomes:\[ \sqrt{2} \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \sin \theta_t \]Solving this, we get:\[ \frac{3}{2} = \sqrt{3} \sin \theta_t \]\[ \sin \theta_t = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]Thus, \( \theta_t = 60^\circ \).

Step 3: Confirm Refracted Angle

From the previous calculations, \( \sin \theta_t = \frac{\sqrt{3}}{2} \), which corresponds to an angle of \( \theta_t = 60^\circ \) in medium 2. Thus, the correct answer is option (b) \( 60^\circ \).

Key Concepts

Refractive Index

In optics, the refractive index is a fundamental property of materials that measures how much the path of light is bent, or refracted, when entering a material. The refractive index is denoted typically as "n" and defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. An important value, it tells us how much slower light travels in a medium compared to vacuum.

• If the refractive index of a medium is greater than one, light travels slower in that medium compared to vacuum.
• For mediums at the boundary, like in our exercise, knowing their refractive indices is crucial to calculate angles of incidence and refraction.

In the original exercise, we have medium 1 with a refractive index of \( \sqrt{2} \) and medium 2 with a refractive index of \( \sqrt{3} \). These values dictate how light behaves when transitioning from one medium to another. By understanding the refractive indices of both mediums, you can predict how much the light will bend upon entering each medium.

Snell's Law

Snell's Law is a formula used to describe the relationship between the angles of incidence and refraction when a wave, such as light, passes through a boundary between two different isotropic media. This law is named after Dutch mathematician Willebrord Snellius. The law states that the ratio of the sine of the angle of incidence (\( \theta_i \)) to the sine of the angle of refraction (\( \theta_t \)) is equivalent to the ratio of the refractive indices of the two media. This is expressed as:\[ n_1 \sin \theta_i = n_2 \sin \theta_t\]

• \( n_1 \) and \( n_2 \) are the refractive indices of the first and second medium respectively.
• \( \theta_i \) is the angle of incidence, while \( \theta_t \) is the angle of refraction.

In the exercise solution, applying Snell's Law allowed us to calculate that \( \theta_t = 60^\circ \). Once you know the angle of incidence and the refractive indices, you can easily calculate the angle of refraction using this reliable law.

Direction Cosine

Direction cosines are the cosines of the angles formed by a vector with the coordinate axes and are essential in vector mathematics, especially in physics. They provide insights into the orientation of the vector in space. If a vector \( \mathbf{A} \) is directed in space, the cosines of the angles it forms with the x-axis, y-axis, and z-axis are known as the direction cosines.

• Direction cosines help determine how much of the vector is pointing in the direction of each axis.
• They are calculated by dividing each component of the vector by the vector's magnitude.

In our exercise, a vector \( \mathbf{A} = 6\sqrt{3} \hat{\mathbf{i}} + 8\sqrt{3} \hat{\mathbf{j}} - 10 \hat{\mathbf{k}} \) was given. To find the angle of incidence \( \theta_i \) with the z-axis, we used the direction cosine relative to that axis. This established the initial angle needed to apply Snell's Law and find the refracted angle.