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 solve the problem of finding the angle of total internal reflection given the wavelengths of light in air and in a medium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Wavelengths:**
- Wavelength in air (λ_air) = 6000 Å
- Wavelength in medium (λ_medium) = 4000 Å

2. **Calculate the Refractive Index (n) of the Medium:**
- The refractive index (n) is given by the formula:
\[ n = \frac{\lambda_{\text{air}}}{\lambda_{\text{medium}}} \]
- Substituting the values:
\[ n = \frac{6000 \, \text{Å}}{4000 \, \text{Å}} = \frac{6000}{4000} = \frac{3}{2} \]

3. **Understanding Total Internal Reflection:**
- Total internal reflection occurs when light travels from a denser medium (in this case, the medium with n = 3/2) to a rarer medium (air, where n = 1).
- The critical angle (θ_c) is the angle of incidence at which the angle of refraction is 90 degrees.

4. **Apply Snell's Law:**
- According to Snell's Law:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
- For total internal reflection:
- \( n_1 = \frac{3}{2} \) (refractive index of the medium)
- \( n_2 = 1 \) (refractive index of air)
- \( \theta_2 = 90^\circ \)
- Thus, we have:
\[ \frac{3}{2} \sin(\theta_c) = 1 \cdot \sin(90^\circ) \]
- Since \(\sin(90^\circ) = 1\), we can simplify this to:
\[ \frac{3}{2} \sin(\theta_c) = 1 \]

5. **Solve for the Critical Angle (θ_c):**
- Rearranging gives:
\[ \sin(\theta_c) = \frac{2}{3} \]
- Now, to find the critical angle:
\[ \theta_c = \sin^{-1}\left(\frac{2}{3}\right) \]

6. **Conclusion:**
- The angle of total internal reflection is:
\[ \theta_c = \sin^{-1}\left(\frac{2}{3}\right) \]