An optical fibre having core of refractive index `sqrt3` and cladding of refractive index 1.5 is kept in air. The maximum angle of acceptance is

Options

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

Answer

To find the maximum angle of acceptance (θA) for the optical fiber, we can use the formula:

\[ \sin(\theta_A) = \sqrt{n_1^2 - n_2^2} \]

where:
- \( n_1 \) is the refractive index of the core,
- \( n_2 \) is the refractive index of the cladding.

### Step 1: Identify the given values
- The refractive index of the core, \( n_1 = \sqrt{3} \)
- The refractive index of the cladding, \( n_2 = 1.5 \)

### Step 2: Square the refractive indices
Calculate \( n_1^2 \) and \( n_2^2 \):
- \( n_1^2 = (\sqrt{3})^2 = 3 \)
- \( n_2^2 = (1.5)^2 = 2.25 \)

### Step 3: Substitute into the formula
Now substitute these values into the formula for sine:
\[ \sin(\theta_A) = \sqrt{n_1^2 - n_2^2} = \sqrt{3 - 2.25} \]

### Step 4: Calculate the difference
Calculate \( 3 - 2.25 \):
\[ 3 - 2.25 = 0.75 \]

### Step 5: Find the square root
Now find the square root:
\[ \sin(\theta_A) = \sqrt{0.75} \]

### Step 6: Simplify the square root
We can express \( 0.75 \) as:
\[ 0.75 = \frac{3}{4} \quad \text{so} \quad \sqrt{0.75} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Step 7: Find the angle
Now, we need to find \( \theta_A \):
\[ \theta_A = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \]

### Step 8: Determine the angle
From trigonometric values, we know:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \theta_A = 60^\circ \]

### Final Answer
The maximum angle of acceptance is \( 60^\circ \).

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