Question

Read the passage given below and answer the questions that follow:

When a ray of light travelling from a denser medium to a rarer medium is incident at the interface of the two media at an angle greater than the critical angle for the two media, the ray is ‘totally’ reflected back into the denser medium. Since refractive index depends upon the wavelength of light, the critical angle for a given pair of media is different for different wavelengths (colours) of light.

Critical angle for glass-air interface can be written as

`sin i_c = 1/(""_an_g) = "velocity of light in glass"/"velocity of light in air"`

1. Name the principle on the basis of which optical fibres work.
2. Calculate the critical angle for the glass and water pair. Given the refractive index of glass is 1.62 and that of water is 1.33.
3. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is `4/3` (n = 3.14).
4. 
   
   Light rays of blue, green and red colours are incident on a right-angled prism, as shown. Green ray makes a grazing emergence. Out of red and blue light, which one comes out of the prism without suffering total internal reflection?

Answer 1

i. The optical fibre works on the principle of total internal reflection.

ii. `""_wn_g = 1/(sin i_c)`

`n_g/n_w = 1/(sin i_c)`

⇒ `sin i_c = n_w/n_g`

`sin i_c = 1.33/1.62`

`i_c = sin^-1(133/162)`

= 55°

iii.

`sin i_c = 1/n_w = 3/4`

r = h tan i_c

A = πr² = πh² tan² i_c

= `3.14 xx (0.8)^2 (3/sqrt7)^2`

= `3.14 xx 0.64 xx 9/7`

= 2.6 m²

iv. For green light, the critical angle for the prism-air interface is 45°.

As, n = `1/sin i_c` and n_red < n_green

So, `(i_c)_"Red" > (i_c)_"Green"`

⇒ Critical angle for red light > 45°.

While the angle of incidence for red at the prism-air interface is 45°, which is less than its critical angle value, so red will come out of the prism without suffering total internal reflection.