The wavelength of light in two liquids ‘ x ' and ‘ y ' is 3500 Å and 7000 Å, then the critical angle of x relative to y will be

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `15^(@)`

Answer

To find the critical angle of liquid 'x' relative to liquid 'y', we can use the formula for the critical angle in terms of the refractive indices of the two media. The refractive index (n) of a medium can be related to the wavelength of light in that medium (λ) using the formula:

\[ n = \frac{\lambda_0}{\lambda} \]

where \( \lambda_0 \) is the wavelength of light in vacuum (or air), and \( \lambda \) is the wavelength of light in the medium.

### Step-by-Step Solution:

1. **Identify the Wavelengths**:
- Wavelength in liquid 'x' (λ_x) = 3500 Å
- Wavelength in liquid 'y' (λ_y) = 7000 Å

2. **Calculate the Refractive Indices**:
- Assume the wavelength of light in vacuum (λ_0) = 1 Å (for simplicity, as it will cancel out).
- For liquid 'x':
\[ n_x = \frac{\lambda_0}{\lambda_x} = \frac{1}{3500} \]
- For liquid 'y':
\[ n_y = \frac{\lambda_0}{\lambda_y} = \frac{1}{7000} \]

3. **Using the Critical Angle Formula**:
The critical angle (C) can be calculated using Snell's law:
\[ \sin C = \frac{n_y}{n_x} \]

4. **Substituting the Values**:
\[ \sin C = \frac{n_y}{n_x} = \frac{\frac{1}{7000}}{\frac{1}{3500}} = \frac{3500}{7000} = \frac{1}{2} \]

5. **Finding the Critical Angle**:
To find the angle whose sine is \( \frac{1}{2} \):
\[ C = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ \]

### Final Answer:
The critical angle of liquid 'x' relative to liquid 'y' is \( 30^\circ \).

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