A liquid is placed in a hollow prism, of angle `60^(@)`. If angle of the minimum deviation is `30^(@)`, what is the refractive index of the liquid?

Options

A. 1.41
B. 1.5
C. 1.65
D. 1.95

Answer

To find the refractive index of the liquid in the hollow prism, we can follow these steps:

### Step 1: Understand the Given Information
We have a hollow prism with an angle \( A = 60^\circ \) and the angle of minimum deviation \( D_m = 30^\circ \). We need to find the refractive index \( \mu \) of the liquid inside the prism.

### Step 2: Use the Formula for Minimum Deviation
In the case of minimum deviation, the relationship between the angle of incidence \( i \), angle of refraction \( r \), and the prism angle \( A \) is given by:
\[ D_m = 2i - A \]
From this, we can express the angle of incidence \( i \):
\[ i = \frac{D_m + A}{2} \]

### Step 3: Substitute the Values
Substituting the known values:
\[ i = \frac{30^\circ + 60^\circ}{2} = \frac{90^\circ}{2} = 45^\circ \]

### Step 4: Calculate the Angle of Refraction
For minimum deviation, the angle of refraction \( r \) is given by:
\[ r = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \]

### Step 5: Apply Snell's Law
According to Snell's law:
\[ n_1 \sin i = \mu \sin r \]
Where \( n_1 \) is the refractive index of air, which is approximately 1. Thus:
\[ \sin i = \mu \sin r \]
Substituting the values of \( i \) and \( r \):
\[ 1 \cdot \sin(45^\circ) = \mu \cdot \sin(30^\circ) \]

### Step 6: Substitute the Sine Values
Using the known sine values:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sin(30^\circ) = \frac{1}{2} \]
We get:
\[ \frac{1}{\sqrt{2}} = \mu \cdot \frac{1}{2} \]

### Step 7: Solve for the Refractive Index \( \mu \)
Rearranging the equation gives:
\[ \mu = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 8: Approximate the Value
Calculating \( \sqrt{2} \):
\[ \mu \approx 1.41 \]

### Conclusion
The refractive index of the liquid is approximately \( 1.41 \).

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