Question

When a parallel beam of light enters water surface obliquely at some angle, what is the effect on the width of the beam ?
(b) With the help of a ray diagram, show that a straw appears bent when it is partly dipped in water and explain it.
(c) Explain the transmission of optical signal through an optical fibre by a diagram.

Hint:

(a) Beam width increases, w' = w cos r / cos i. (b) Rays from the dipped part bend away from the normal in air, so the part looks raised and the straw seems bent. (c) Total internal reflection at the core-cladding boundary carries the signal.

Answer 1

Step 1: Width of the beam (part a):
Let a parallel beam of width \(w\) fall on the water surface at angle of incidence \(i\). It covers a length \(L\) of the surface, so \(w = L\cos i\).
After refraction the angle of refraction is \(r\), and the width of the refracted beam is \(w' = L\cos r\).
Dividing, \(\dfrac{w'}{w} = \dfrac{\cos r}{\cos i}\). Light going from air to water bends towards the normal, so \(r < i\) and \(\cos r > \cos i\).
So \(w' > w\). On entering water obliquely the width of a parallel beam increases.

Step 2: Straw appears bent (part b):
Ray diagram in words: draw the water surface as a horizontal line and the straw slanting into the water. Take a point O on the dipped part of the straw. Two rays leave O and travel up through the water.
At the surface each ray goes from water (denser) to air (rarer), so it bends away from the normal.
An observer in air traces the bent rays straight back. They seem to come from a point I which is higher than O and nearer the surface. Every point of the dipped part is raised in the same way.
So the dipped part looks lifted while the part in air is seen at its true place. The straw seems bent at the water surface.
Quantitatively, apparent depth = real depth / n, with n = 4/3 for water. A point at depth d appears at about 3d/4.

Step 3: Optical fibre (part c):
An optical fibre is a thin glass or plastic thread. It has a central core of higher refractive index \(n_1\) and a covering called cladding of lower refractive index \(n_2\).
Diagram in words: draw a long strip (the core) with thin cladding on both sides. Draw a light ray entering one end of the core. The ray meets the core-cladding boundary at an angle larger than the critical angle, reflects, meets the opposite boundary, reflects again and so zig-zags along the fibre to the other end.
The condition is \(\sin\theta > n_2/n_1\), where \(\theta\) is the angle with the normal at the boundary. Then total internal reflection takes place every time and almost no light is lost.
In this way the optical signal (light pulses carrying the information) travels long distances, even when the fibre is bent gently.

Final Answer:
(a) The width of the beam increases (\(w' = w\cos r/\cos i\)). (b) The dipped part appears raised because rays from it bend away from the normal in air, so the straw looks bent. (c) Light travels along the fibre by repeated total internal reflection at the core-cladding boundary.