A glass prism of angle `A = 60^(@)` gives minimum angle of deviation `theta ~~ 30^(@)` with the maximum error of `1^(@)` when a beam of parallel light passed through the prism during an experiment. Find the permissible error in the measurement of refractive index `mu` of the material of the prism.

Explanation

To find the permissible error in the measurement of the refractive index \( \mu \) of the material of the prism, we will follow these steps:

### Step 1: Identify Given Values
- Angle of the prism \( A = 60^\circ \)
- Minimum angle of deviation \( \theta = 30^\circ \)
- Maximum error in angle of deviation \( \Delta \theta = 1^\circ \)

### Step 2: Calculate the Angle of Incidence
The angle of incidence \( i \) can be calculated using the formula:
\[ i = \frac{A + \theta}{2} \]
Substituting the values:
\[ i = \frac{60^\circ + 30^\circ}{2} = \frac{90^\circ}{2} = 45^\circ \]

### Step 3: Use the Formula for Refractive Index
The refractive index \( \mu \) can be calculated using the formula:
\[ \sin i = \mu \sin \left(\frac{A}{2}\right) \]
Substituting \( i = 45^\circ \) and \( A = 60^\circ \):
\[ \sin(45^\circ) = \mu \sin(30^\circ) \]
We know that \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \) and \( \sin(30^\circ) = \frac{1}{2} \):
\[ \frac{1}{\sqrt{2}} = \mu \cdot \frac{1}{2} \]
Solving for \( \mu \):
\[ \mu = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 4: Differentiate to Find \( \Delta \mu \)
To find the permissible error in the refractive index \( \Delta \mu \), we differentiate the equation:
\[ \mu = \frac{\sin(i)}{\sin\left(\frac{A}{2}\right)} \]
Using the chain rule, we have:
\[ \Delta \mu = \csc\left(\frac{A}{2}\right) \cdot \cos\left(i\right) \cdot \frac{\Delta \theta}{2} \]
Substituting \( A = 60^\circ \) and \( i = 45^\circ \):
\[ \Delta \mu = \csc(30^\circ) \cdot \cos(45^\circ) \cdot \frac{\Delta \theta}{2} \]
We know that \( \csc(30^\circ) = 2 \) and \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \Delta \mu = 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1^\circ}{2} \]
Converting \( 1^\circ \) to radians:
\[ 1^\circ = \frac{\pi}{180} \]
Thus,
\[ \Delta \mu = 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi/180}{2} = \frac{\pi}{180\sqrt{2}} \]

### Step 5: Calculate the Percentage Error
The percentage error in the measurement of refractive index \( \mu \) is given by:
\[ \text{Percentage Error} = \frac{\Delta \mu}{\mu} \times 100 \]
Substituting \( \mu = \sqrt{2} \):
\[ \text{Percentage Error} = \frac{\frac{\pi}{180\sqrt{2}}}{\sqrt{2}} \times 100 = \frac{\pi}{180} \times 100 \]
This simplifies to:
\[ \text{Percentage Error} = \frac{100\pi}{180} = \frac{5\pi}{9} \% \]

### Final Answer
The permissible error in the measurement of the refractive index \( \mu \) of the material of the prism is:
\[ \frac{5\pi}{9} \% \text{ (approximately 17.5\%)} \]