For a light pipe as shown in the figure :(A) Optical density of core…

The correct answer is Option (3) → (A), (C) only

A light pipe guides light by total internal reflection at the core-cladding boundary. Let \( n_1 \) be the refractive index of the core and \( n_2 \) that of the cladding.

At the entry face (air to core), Snell's law gives

    \( \displaystyle \sin i = n_1 \sin r \;\Rightarrow\; n_1 = \frac{\sin i}{\sin r} \)

The angle \( r \) is measured from the axis (the normal to the entry face) and \( \theta \) from the normal to the side face, which is perpendicular to the axis. So

    \( \displaystyle \theta = 90^\circ - r \)

For the ray to be just totally reflected, \( \theta \) is the critical angle at the core-cladding boundary:

    \( \displaystyle n_1 \sin\theta = n_2 \;\Rightarrow\; n_2 = \frac{\sin\theta\,\sin i}{\sin r} \)

(A) Optical density of core should be greater than the optical density of cladding — Correct. Total internal reflection happens only when light goes from a denser to a rarer medium.

(B) r and \( \theta \) will always be equal — Incorrect. Since \( \theta = 90^\circ - r \), they are equal only when \( r = 45^\circ \).

(C) Optical density of cladding is \( \sin\theta.\sin i/\sin r \) — Correct. This is \( n_2 = n_1\sin\theta \) with \( n_1 = \sin i/\sin r \).

(D) Optical density of cladding is \( \sin r.\sin i/\sin\theta \) — Incorrect. The roles of \( \sin r \) and \( \sin\theta \) are interchanged.

NCERT (2026-27):
Total Internal Reflection in Nature and Its Technological Applications – “… Optical fibres too make use of the phenomenon of total internal reflection. Optical fibres are fabricated with high quality composite glass/quartz fibres. Each fibre consists of a core and cladding. The refractive index of the material of the core is higher than that of the cladding.”