Total Internal Reflection: Critical Angle Derivation

TITLE: Total Internal Reflection: Critical Angle Derivation DESCRIPTION: Understand Total Internal Reflection (TIR) and derive the critical angle using Snell's Law. Essential for JEE Physics.

Concept Overview

This question tests the understanding of Total Internal Reflection (TIR), a phenomenon crucial in ray optics. We will derive the condition for TIR and the formula for the critical angle by applying Snell's Law at the interface between two media. This involves understanding how light bends when passing from a denser to a rarer medium and the specific angle at which it ceases to refract.

Solution:

Step 1: State Snell's Law. Snell's Law describes the relationship between the angles of incidence and refraction and the refractive indices of two media. It is given by: $n_1 \sin \theta_1 = n_2 \sin \theta_2$ Here, $n_1$ is the refractive index of the first medium, $\theta_1$ is the angle of incidence, $n_2$ is the refractive index of the second medium, and $\theta_2$ is the angle of refraction.

Step 2: Define the conditions for Total Internal Reflection. TIR occurs when light travels from a medium of higher refractive index ($n_1$) to a medium of lower refractive index ($n_2$). This means $n_1 > n_2$. As the angle of incidence ($\theta_1$) increases, the angle of refraction ($\theta_2$) also increases.

Step 3: Define the critical angle. The critical angle, denoted by $\theta_c$, is the specific angle of incidence in the denser medium for which the angle of refraction in the rarer medium is exactly $90^\circ$. At this point, the refracted ray travels along the interface between the two media.

Step 4: Apply Snell's Law for the critical angle condition. When the angle of incidence is equal to the critical angle ($\theta_1 = \theta_c$), the angle of refraction is $90^\circ$ ($\theta_2 = 90^\circ$). Substituting these values into Snell's Law: $n_1 \sin \theta_c = n_2 \sin 90^\circ$

Step 5: Simplify the equation using $\sin 90^\circ = 1$. Since $\sin 90^\circ$ is equal to 1, the equation becomes: $n_1 \sin \theta_c = n_2 (1)$ $n_1 \sin \theta_c = n_2$

Step 6: Derive the formula for the critical angle. To find the critical angle, we rearrange the equation to solve for $\sin \theta_c$: $\sin \theta_c = \frac{n_2}{n_1}$ This formula shows that the sine of the critical angle depends on the ratio of the refractive indices of the two media. For TIR to occur, the angle of incidence must be greater than or equal to the critical angle ($\theta_1 \ge \theta_c$).

Step 7: Consider the special case of light going from glass to air. A common scenario is light traveling from glass ($n_1 \approx 1.5$) to air ($n_2 \approx 1.0$). In this case, the critical angle is: $\sin \theta_c = \frac{1.0}{1.5} = \frac{2}{3}$ $\theta_c = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.8^\circ$ This means that if light strikes the glass-air interface at an angle greater than approximately $41.8^\circ$, it will be totally internally reflected back into the glass.

Key Takeaways:

• Total Internal Reflection (TIR) occurs when light travels from a denser medium ($n_1$) to a rarer medium ($n_2$), with $n_1 > n_2$.
• The critical angle ($\theta_c$) is the angle of incidence for which the angle of refraction is $90^\circ$.
• The relationship between the critical angle and refractive indices is given by $\sin \theta_c = \frac{n_2}{n_1}$.
• TIR happens when the angle of incidence is greater than or equal to the critical angle ($\theta_1 \ge \theta_c$).

Answer: $\sin \theta_c = \frac{n_2}{n_1}$