JEE Numerical Problems on Total Internal Reflection

1. A light ray is incident on a glass-air interface at an angle of 45 degrees. What is the angle of refraction?

Solution:

Using Snell’s law, we have:

$$n_1\sin\theta_1 = n_2\sin\theta_2$$

Where:

• $$n_1$$ is the refractive index of the first medium (glass)
• $$\theta_1$$ is the angle of incidence
• $$n_2$$ is the refractive index of the second medium (air)
• $$\theta_2$$ is the angle of refraction

Substituting the given values, we get:

$$1.5\sin45^\circ = 1\sin\theta_2$$

Solving for $$\theta_2$$, we get:

$$\theta_2 = \sin^{-1}(1.5\sin45^\circ) = 29.46^\circ$$

Therefore, the angle of refraction is 29.46 degrees.

2. A light ray is traveling from a medium with refractive index 1.5 to a medium with refractive index 1.0. What is the critical angle for total internal reflection?

Solution:

The critical angle for total internal reflection is given by:

$$\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)$$

Where:

• $$\theta_c$$ is the critical angle
• $$n_1$$ is the refractive index of the first medium
• $$n_2$$ is the refractive index of the second medium

Substituting the given values, we get:

$$\theta_c = \sin^{-1}\left(\frac{1.0}{1.5}\right) = 41.81^\circ$$

Therefore, the critical angle for total internal reflection is 41.81 degrees.

3. A glass prism has a refractive index of 1.5. What is the minimum angle of incidence for a light ray to undergo total internal reflection inside the prism?

Solution:

The minimum angle of incidence for total internal reflection inside a prism is given by:

$$\theta_i = \sin^{-1}\left(\frac{n_2}{n_1}\right)$$

Where:

• $$\theta_i$$ is the minimum angle of incidence
• $$n_1$$ is the refractive index of the first medium (air)
• $$n_2$$ is the refractive index of the second medium (glass)

Substituting the given values, we get:

$$\theta_i = \sin^{-1}\left(\frac{1.5}{1}\right) = 41.81^\circ$$

Therefore, the minimum angle of incidence for a light ray to undergo total internal reflection inside the prism is 41.81 degrees.

4. A light ray is incident on a plane mirror at an angle of 30 degrees. What is the angle of reflection?

Solution:

The angle of reflection for a light ray incident on a plane mirror is equal to the angle of incidence.

Therefore, the angle of reflection is 30 degrees.

5. A light ray is traveling from a medium with refractive index 1.0 to a medium with refractive index 1.5. What is the angle of incidence for which the light ray will be refracted at an angle of 30 degrees?

Solution:

Using Snell’s law, we have:

$$n_1\sin\theta_1 = n_2\sin\theta_2$$

Where:

• $$n_1$$ is the refractive index of the first medium (air)
• $$\theta_1$$ is the angle of incidence
• $$n_2$$ is the refractive index of the second medium (glass)
• $$\theta_2$$ is the angle of refraction

Substituting the given values, we get:

$$1\sin\theta_1 = 1.5\sin30^\circ$$

Solving for $$\theta_1$$, we get:

$$\theta_1 = \sin^{-1}(1.5\sin30^\circ) = 19.47^\circ$$

Therefore, the angle of incidence for which the light ray will be refracted at an angle of 30 degrees is 19.47 degrees.

CBSE Board Exam Numerical Problems on Total Internal Reflection

1. A light ray is incident on a glass-air interface at an angle of 45 degrees. What is the angle of refraction?

Solution:

Using Snell’s law, we have:

$$n_1\sin\theta_1 = n_2\sin\theta_2$$

Where:

• $$n_1$$ is the refractive index of the first medium (glass)
• $$\theta_1$$ is the angle of incidence
• $$n_2$$ is the refractive index of the second medium (air)
• $$\theta_2$$ is the angle of refraction

Substituting the given values, we get:

$$1.5\sin45^\circ = 1\sin\theta_2$$

Solving for $$\theta_2$$, we get:

$$\theta_2 = \sin^{-1}(1.5\sin45^\circ) = 29.46^\circ$$

Therefore, the angle of refraction is 29.46 degrees.

2. A light ray is traveling from a medium with refractive index 1.5 to a medium with refractive index 1.0. What is the critical angle for total internal reflection?

Solution:

The critical angle for total internal reflection is given by:

$$\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)$$

Where:

• $$\theta_c$$ is the critical angle
• $$n_1$$ is the refractive index of the first medium
• $$n_2$$ is the refractive index of the second medium

Substituting the given values, we get:

$$\theta_c = \sin^{-1}\left(\frac{1.0}{1.5}\right) = 41.81^\circ$$

Therefore, the critical angle for total internal reflection is 41.81 degrees.

3. A glass prism has a refractive index of 1.5. What is the minimum angle of incidence for a light ray to undergo total internal reflection inside the prism?

Solution:

The minimum angle of incidence for total internal reflection inside a prism is given by:

$$\theta_i = \sin^{-1}\left(\frac{n_2}{n_1}\right)$$

Where:

• $$\theta_i$$ is the minimum angle of incidence
• $$n_1$$ is the refractive index of the first medium (air)
• $$n_2$$ is the refractive index of the second medium (glass)

Substituting the given values, we get:

$$\theta_i = \sin^{-1}\left(\frac{1.5}{1}\right) = 41.81^\circ$$

Therefore, the minimum angle of incidence for a light ray to undergo total internal reflection inside the prism is 41.81 degrees.

4. A light ray is incident on a plane mirror at an angle of 30 degrees. What is the angle of reflection?

Solution:

The angle of reflection for a light ray incident on a plane mirror is equal to the angle of incidence.

Therefore, the angle of reflection is 30 degrees.

5. A light ray is traveling from a medium with refractive index 1.0 to a medium with refractive index 1.5. What is the angle of incidence for which the light ray will be refracted at an angle of 30 degrees?

Solution:

Using Snell’s law, we have:

$$n_1\sin\theta_1 = n_2\sin\theta_2$$

Where:

• $$n_1$$ is the refractive index of the first medium (air)
• $$\theta_1$$ is the angle of incidence
• $$n_2$$ is the refractive index of the second medium (glass)
• $$\theta_2$$ is the angle of refraction

Substituting the given values, we get:

$$1\sin\theta_1 = 1.5\sin30^\circ$$

Solving for $$\theta_1$$, we get:

$$\theta_1 = \sin^{-1}(1.5\sin30^\circ) = 19.47^\circ$$

Therefore, the angle of incidence for which the light ray will be refracted at an angle of 30 degrees is 19.47 degrees.