An optical fibre has numerical aperture (NA) of 0.3 and refractive index $\eta_2$ of cladding material is 1.6. What is the refractive index of core material?
1.63
To determine the refractive index of the core material, we will use the formula for the numerical aperture (NA) of an optical fibre, which is given by:
\(NA = \sqrt{n_1^2 - n_2^2}\)
where \(n_1\) is the refractive index of the core and \(n_2\) is the refractive index of the cladding.
We are given:
- Numerical Aperture, \(NA = 0.3\)
- Refractive index of the cladding, \(n_2 = 1.6\)
We need to find \(n_1\). Plugging the values into the formula:
\(0.3 = \sqrt{n_1^2 - (1.6)^2}\)
Squaring both sides, we get:
\(0.09 = n_1^2 - 2.56\)
Rearranging the equation to solve for \(n_1^2\):
\(n_1^2 = 0.09 + 2.56\)
\(n_1^2 = 2.65\)
Taking the square root of both sides to find \(n_1\):
\(n_1 = \sqrt{2.65}\)
\(n_1 \approx 1.63\)
Therefore, the refractive index of the core material \((n_1)\) is approximately 1.63.
The correct answer is 1.63.
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