Find the angle of refraction in a medium `(mu=2)` if light is incident in vacuum, making an angle equal to twice the critical angle.

Options

A. `sin^(-1)((sqrt3)/4)`
B. `sin^(-1)((sqrt3)/2)`
C. `sin^(-1)((1)/2)`
D. `sin^(-1)(2)`

Explanation

To find the angle of refraction in a medium with a refractive index of \( \mu = 2 \) when light is incident from a vacuum at an angle equal to twice the critical angle, we can follow these steps:

### Step 1: Calculate the Critical Angle
The critical angle \( \theta_c \) can be found using Snell's law at the boundary between two media, where the light goes from a medium with a higher refractive index to a lower one. The formula is:

\[ \mu_1 \sin(\theta_c) = \mu_2 \sin(90^\circ) \]

Here, \( \mu_1 = 2 \) (the refractive index of the medium) and \( \mu_2 = 1 \) (the refractive index of vacuum). Therefore, we have:

\[ 2 \sin(\theta_c) = 1 \]

This simplifies to:

\[ \sin(\theta_c) = \frac{1}{2} \]

Taking the inverse sine gives us:

\[ \theta_c = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ \]

### Step 2: Determine the Incident Angle
The problem states that the incident angle \( \theta_1 \) is twice the critical angle:

\[ \theta_1 = 2 \times \theta_c = 2 \times 30^\circ = 60^\circ \]

### Step 3: Apply Snell's Law
Now we apply Snell's law to find the angle of refraction \( \theta_2 \):

\[ \mu_1 \sin(\theta_1) = \mu_2 \sin(\theta_2) \]

Substituting the known values:

\[ 2 \sin(60^\circ) = 1 \sin(\theta_2) \]

Calculating \( \sin(60^\circ) \):

\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

Thus, we have:

\[ 2 \cdot \frac{\sqrt{3}}{2} = \sin(\theta_2) \]

This simplifies to:

\[ \sqrt{3} = \sin(\theta_2) \]

### Step 4: Find the Angle of Refraction
To find \( \theta_2 \), we take the inverse sine:

\[ \theta_2 = \sin^{-1}(\sqrt{3}) \]

However, since \( \sqrt{3} \) is greater than 1, this means that the angle of refraction is not possible in this case, indicating total internal reflection occurs.

### Conclusion
Thus, the angle of refraction cannot be determined as light does not refract into the medium when the incident angle is greater than the critical angle.

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