Wavelength of given light waves in air and in a medium are 6000 Å and 3000 Å, respectively. The critical angle is

Options

A. `tan^(-1)(2/3)`
B. `tan^(-1)(3/2)`
C. `sin^(-1)(2)`
D. `sin^(-1)(3/2)`

Explanation

To find the critical angle for the given light waves in air and in a medium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Wavelength in air (λ_air) = 6000 Å
- Wavelength in medium (λ_medium) = 3000 Å

2. **Calculate the refractive index (μ) of the medium:**
The refractive index (μ) can be calculated using the formula:
\[ \mu = \frac{\text{Wavelength in air}}{\text{Wavelength in medium}} = \frac{\lambda_{\text{air}}}{\lambda_{\text{medium}}} \]
Substituting the values:
\[ \mu = \frac{6000 \, \text{Å}}{3000 \, \text{Å}} = 2 \]

3. **Use the formula for the critical angle (θ_c):**
The critical angle can be found using the formula:
\[ \sin(\theta_c) = \frac{1}{\mu} \]
Substituting the value of μ:
\[ \sin(\theta_c) = \frac{1}{2} \]

4. **Calculate the critical angle (θ_c):**
To find the critical angle, take the inverse sine:
\[ \theta_c = \sin^{-1}\left(\frac{1}{2}\right) \]
The value of \(\sin^{-1}\left(\frac{1}{2}\right)\) is:
\[ \theta_c = 30^\circ \]

### Final Answer:
The critical angle is \(30^\circ\).

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