The critical angle of a medium with respect to air is `45^(@)` . The refractive index of medium is

Options

A. 1.41
B. 1.2
C. 1.5
D. 2

Answer

To find the refractive index of a medium with respect to air given that the critical angle is \( 45^\circ \), we can use Snell's Law, which relates the angles of incidence and refraction to the refractive indices of the two media.

### Step-by-Step Solution:

1. **Understanding Critical Angle**: The critical angle (\( \theta_c \)) is the angle of incidence in the denser medium (in this case, the medium) for which the angle of refraction in the less dense medium (air) is \( 90^\circ \).

2. **Using Snell's Law**: According to Snell's Law:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
Where:
- \( n_1 \) is the refractive index of the denser medium (the medium we are looking for).
- \( \theta_1 \) is the angle of incidence (which is the critical angle \( \theta_c \)).
- \( n_2 \) is the refractive index of the less dense medium (air), which is approximately \( 1 \).
- \( \theta_2 \) is the angle of refraction (which is \( 90^\circ \) at the critical angle).

3. **Setting Up the Equation**: Since the angle of refraction is \( 90^\circ \), we can write:
\[ n_1 \sin(45^\circ) = n_2 \sin(90^\circ) \]
Substituting \( n_2 = 1 \) (for air) and \( \sin(90^\circ) = 1 \):
\[ n_1 \sin(45^\circ) = 1 \cdot 1 \]

4. **Calculating \( \sin(45^\circ) \)**: We know that:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]

5. **Substituting the Value**: Now substituting \( \sin(45^\circ) \) into the equation:
\[ n_1 \cdot \frac{1}{\sqrt{2}} = 1 \]

6. **Solving for \( n_1 \)**: Rearranging the equation to find \( n_1 \):
\[ n_1 = \sqrt{2} \]

7. **Numerical Value**: The numerical value of \( \sqrt{2} \) is approximately \( 1.414 \).

### Final Answer:
The refractive index of the medium is \( \sqrt{2} \) or approximately \( 1.414 \).

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