If the fractional difference between the core and cladding refractive indices of the optical fibre is 0.0135 and numerical aperture NA is 0.2425, calculate the refractive indices of core and cladding

Given:
Fractional difference, $\Delta = 0.0135$
Numerical Aperture, $NA = 0.2425$

Formulas:

1. Numerical Aperture: $NA = n_1 \sqrt{2\Delta}$
2. Fractional difference: $\Delta = \frac{n_1 - n_2}{n_1}$ Here, $n_1$ is the refractive index of the core and $n_2$ is the refractive index of the cladding.

Using the numerical aperture formula:

$$NA = n_1 \sqrt{2\Delta}$$

Rearrange to solve for $n_1$:

$$n_1 = \frac{NA}{\sqrt{2\Delta}}$$

Substitute the given values:

$$n_1 = \frac{0.2425}{\sqrt{2 \times 0.0135}}$$ $$n_1 = \frac{0.2425}{\sqrt{0.027}}$$ $$n_1 = \frac{0.2425}{0.164316767}$$ $$n_1 \approx 1.4758$$

Using the fractional difference formula:

$$\Delta = \frac{n_1 - n_2}{n_1}$$

Rearrange to solve for $n_2$:

$$n_1 \Delta = n_1 - n_2$$ $$n_2 = n_1 - n_1 \Delta$$

Factor out $n_1$:

$$n_2 = n_1 (1 - \Delta)$$

Substitute the calculated $n_1$ and the given $\Delta$:

$$n_2 = 1.4758 (1 - 0.0135)$$ $$n_2 = 1.4758 (0.9865)$$ $$n_2 \approx 1.4558$$

The refractive index of the core material is approximately 1.4758.
The refractive index of the cladding material is approximately 1.4558.

The refractive index of the core material is $\boxed{\text{1.4758}}$ and the refractive index of the cladding material is $\boxed{\text{1.4558}}$.