Animated Solution for Physics - Optics: Comprehension Passage

Imagine an optical fiber as a microscopic tunnel for light. It consists of a central core with a higher refractive index $n_1$, surrounded by a cladding with a lower refractive index $n_2$. When light enters the core from an external medium (like air or water) with refractive index $n_0$, it bends towards the normal.

To keep the light trapped inside the core, it must strike the core-cladding interface at an angle greater than or equal to the critical angle ${\theta }_c$. This requirement restricts the angle at which light can initially enter the fiber. The sine of this maximum acceptance angle $i_m$ is known as the Numerical Aperture (NA).

Let's derive a direct formula for the Numerical Aperture. We start by applying Snell's law at the entrance face:

Inside the core, the geometry of the right-angled triangle tells us that the angle of incidence at the core-cladding interface is $90^{\circ }-\theta$. For Total Internal Reflection (TIR) to occur, this angle must be at least ${\theta }_c$:

To find the maximum acceptance angle $i_m$, we set $\theta$ to its maximum value, $90^{\circ }-{\theta }_c$. Substituting this back into our Snell's law equation yields:

We know from the definition of the critical angle that $\sin {\theta }_c=\frac{n_2}{n_1}$. Using the fundamental trigonometric identity $\cos {\theta }_c=\sqrt{1-{\sin }^2{\theta }_c}$, we can express the Numerical Aperture entirely in terms of the refractive indices:

The term $\sqrt{n_1^2-n_2^2}$ is an intrinsic property of the fiber itself, representing its inherent light-gathering capability.

Let's calculate this intrinsic property for the two structures provided in the problem.

For structure $S_1$:

For structure $S_2$:

Now, we systematically evaluate the given options by dividing these intrinsic values by the refractive index of the surrounding medium $n_0$.

Testing Option (a): For $S_1$ in water ($n_0=4/3$): For $S_2$ in the specified liquid ($n_0=\frac{16}{3\sqrt{15}}$): Since both yield $\frac{9}{16}$, Option (a) is correct.

Testing Option (c): For $S_1$ in air ($n_0=1$): For $S_2$ in the specified liquid ($n_0=\frac{4}{\sqrt{15}}$): Since both yield $\frac{3}{4}$, Option (c) is also correct.

Imagine joining these two optical fibers end-to-end longitudinally. For a light ray to successfully travel through the entire combined structure, it must satisfy the Total Internal Reflection conditions of both individual fibers.

If a ray enters at an angle that is accepted by the fiber with the larger NA but exceeds the acceptance angle of the fiber with the smaller NA, it will escape into the cladding as soon as it reaches the second fiber. Therefore, the overall acceptance angle is strictly limited by the fiber with the smaller Numerical Aperture.

Mathematically, the combined Numerical Aperture is simply the minimum of the two:

Since the problem explicitly states that $N{A}_2<N{A}_1$, the combined Numerical Aperture is $N{A}_2$. Thus, Option (d) is the correct answer.