The wavelengths of given light waves in air and in a medium are 6000 A and 4000 A, respectively. The angle of total internal reflection is .

Options

A. `sin^(-1)(2/3)`
B. `sin^(-1)(3/4)`
C. `sin^(-1)(8/9)`
D. `sin^(-1)(1/2)`

Answer

To find the angle of total internal reflection, we will follow these steps:

### Step 1: Determine the Refractive Index
The refractive index (n) of a medium can be calculated using the formula:
\[ n = \frac{\lambda_{\text{air}}}{\lambda_{\text{medium}}} \]
Given:
- Wavelength in air, \(\lambda_{\text{air}} = 6000 \, \text{Å}\)
- Wavelength in medium, \(\lambda_{\text{medium}} = 4000 \, \text{Å}\)

Substituting the values:
\[ n = \frac{6000 \, \text{Å}}{4000 \, \text{Å}} = \frac{6}{4} = \frac{3}{2} = 1.5 \]

### Step 2: Understand Total Internal Reflection
Total internal reflection occurs when light travels from a denser medium to a rarer medium. The critical angle (\(\theta_c\)) is the angle of incidence at which light is refracted at an angle of 90 degrees in the rarer medium.

### Step 3: Apply Snell's Law
Using Snell's law:
\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]
For total internal reflection, we set:
- \(n_1 = n\) (refractive index of the medium, which is \(1.5\))
- \(n_2 = 1\) (refractive index of air)
- \(\theta_2 = 90^\circ\) (since at the critical angle, the refracted angle is 90 degrees)

Thus, we have:
\[ n \sin \theta_c = 1 \cdot \sin 90^\circ \]
This simplifies to:
\[ n \sin \theta_c = 1 \]

### Step 4: Solve for the Critical Angle
Substituting the value of \(n\):
\[ 1.5 \sin \theta_c = 1 \]
\[ \sin \theta_c = \frac{1}{1.5} = \frac{2}{3} \]

Now, to find \(\theta_c\):
\[ \theta_c = \sin^{-1} \left(\frac{2}{3}\right) \]

### Final Answer
The angle of total internal reflection is:
\[ \theta_c = \sin^{-1} \left(\frac{2}{3}\right) \]