GATE 2023 IN Q.45: solution and answer

Official answer: 0.620 to 0.640Source: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Identify the new cladding. Once the glass cladding is removed and the core is dipped in water, the water surrounds the core and does the cladding's job. So $n_1=1.47$ (core) and $n_2=1.33$ (water). The original cladding index $1.44$ no longer plays any part.

2. Numerical aperture of a step-index guide. Light is launched from air, so $n_0=1$: $$\text{NA}=n_0\sin {\theta }_a=\sqrt{n_1^2-n_2^2}$$

3. Substitute: $$n_1^2=(1.47{)}^2=2.1609,n_2^2=(1.33{)}^2=1.7689$$

4. $$n_1^2-n_2^2=2.1609-1.7689=0.3920$$

5. $$\text{NA}=\sqrt{0.3920}\approx 0.62610\Rightarrow \text{NA}\approx 0.626$$ This is less than $1$, so it is a valid acceptance-angle result: ${\theta }_a={\sin }^{-1}(0.626)\approx 38.8^{\circ }$.

6. Sanity check: the original fiber had $\text{NA}=\sqrt{1.47^2-1.44^2}=\sqrt{0.0873}\approx 0.295$. Water has a much lower index than the glass cladding, so the core-to-surround index contrast is larger and the NA should rise. It does: $0.295\to 0.626$.