Total Internal Reflection: Calculating Critical Angle (Medium A to B)

This problem requires us to find the critical angle for a light ray moving from a transparent medium A to another transparent medium B. We are given the refractive index of medium A and the speed of light in medium B, along with the speed of light in vacuum. This involves applying the principles of refraction and total internal reflection.

Physics Principles: Refraction and Critical Angle

Key concepts to understand are:

• Refractive Index ($n$): This property measures how much light slows down when entering a medium. It's calculated as the ratio of the speed of light in vacuum ($c$) to the speed of light in the medium ($v$). The formula is: $ n = \frac{c}{v} $ A higher refractive index means light travels slower in that medium.
• Snell's Law: This law describes how light bends when crossing the boundary between two different media. It states: $ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) $ where $n_1$ and $n_2$ are the refractive indices of the first and second medium, respectively, and $ \theta_1 $ and $ \theta_2 $ are the angles of incidence and refraction, measured from the normal to the boundary.
• Critical Angle ($ \theta_c $): This is the specific angle of incidence in the denser medium for which the angle of refraction in the rarer medium is $ 90^\circ $. Total internal reflection occurs when the angle of incidence is greater than the critical angle. The formula derived from Snell's Law for the critical angle is: $ \sin(\theta_c) = \frac{n_{rarer}}{n_{denser}} $ Note that total internal reflection can only happen when light travels from a denser medium (higher $n$) to a rarer medium (lower $n$).

Step-by-Step Calculation of Critical Angle

Let's calculate the critical angle using the provided information.

Calculating Refractive Index of Medium B

We are given:

• Speed of light in medium B, $v_B = 2.4 \times 10^8 \text{ m/s}$
• Speed of light in vacuum, $c = 3.0 \times 10^8 \text{ m/s}$

Using the refractive index formula:

$ n_B = \frac{c}{v_B} = \frac{3.0 \times 10^8 \text{ m/s}}{2.4 \times 10^8 \text{ m/s}} $ $ n_B = \frac{3.0}{2.4} = \frac{30}{24} = \frac{5}{4} $ $ n_B = 1.25 $

Identifying Denser and Rarer Mediums

We compare the refractive indices of the two media:

• Refractive index of medium A, $n_A = 1.6$
• Calculated refractive index of medium B, $n_B = 1.25$

Since $ n_A (1.6) > n_B (1.25) $, medium A is optically denser, and medium B is optically rarer. This is the necessary condition for total internal reflection when light travels from A to B.

Calculating Critical Angle $ \theta_c $

Now, we apply the critical angle formula using $ n_A $ as the denser medium's index and $ n_B $ as the rarer medium's index:

$ \sin(\theta_c) = \frac{n_{rarer}}{n_{denser}} = \frac{n_B}{n_A} $ $ \sin(\theta_c) = \frac{1.25}{1.6} $

To find the value, we can convert the fraction to a decimal:

$ \sin(\theta_c) = \frac{1.25}{1.60} = \frac{125}{160} $

Simplifying the fraction (e.g., dividing numerator and denominator by 5 twice, or recognizing $1.6 = 16/10$):

$ \sin(\theta_c) = \frac{1.25}{1.6} = 0.78125 $

The critical angle $ \theta_c $ is the angle whose sine is $0.78125$. Therefore:

$ \theta_c = \sin^{-1}(0.78125) $

Final Critical Angle Result

Based on the calculations, the sine of the critical angle for light traveling from medium A to medium B is $0.78125$. The critical angle is thus represented as $ \sin^{-1}(0.78125) $. This matches one of the provided options.