The refractive index of medium with respect to air, for the critical angle `60^(@)` is

Options

A. 1.6
B. 1.5
C. 1.2
D. 1.15

Answer

To find the refractive index of a medium with respect to air for a critical angle of \(60^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Critical Angle**: The critical angle (\(C\)) is defined as the angle of incidence in a denser medium at which the angle of refraction in the rarer medium is \(90^\circ\). For this problem, the critical angle is given as \(60^\circ\).

2. **Applying Snell's Law**: According to Snell's Law, we have:
\[ n_1 \sin(i) = n_2 \sin(r) \]
where:
- \(n_1\) is the refractive index of the denser medium (the medium we are trying to find).
- \(n_2\) is the refractive index of the rarer medium (air in this case, which is approximately \(1\)).
- \(i\) is the angle of incidence (which will be the critical angle \(C\)).
- \(r\) is the angle of refraction (which is \(90^\circ\) at the critical angle).

3. **Substituting Values**: Since we know the critical angle is \(60^\circ\) and at this angle, the angle of refraction is \(90^\circ\), we can substitute:
\[ n_1 \sin(60^\circ) = n_2 \sin(90^\circ) \]
Here, \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\) and \(\sin(90^\circ) = 1\).

4. **Rearranging the Equation**: Plugging in the values:
\[ n_1 \cdot \frac{\sqrt{3}}{2} = 1 \cdot 1 \]
This simplifies to:
\[ n_1 \cdot \frac{\sqrt{3}}{2} = 1 \]

5. **Solving for \(n_1\)**: To find \(n_1\), we rearrange the equation:
\[ n_1 = \frac{2}{\sqrt{3}} \]

6. **Calculating the Numerical Value**: Now, calculating the numerical value:
\[ n_1 \approx \frac{2}{1.732} \approx 1.1547 \]
Thus, we can round this to approximately \(1.15\).

### Final Answer:
The refractive index of the medium with respect to air for the critical angle of \(60^\circ\) is approximately \(1.15\).