Refraction involves change in the path of light due to change in the medium.

 

When a beam of light encounters another transparent medium, a part of light gets reflected back into the first medium, while the rest enters the other. The direction of propagation of an obliquely incident ray of light, that enters the other medium, changes at the interface of two media. This phenomenon is called refraction of light. 

(i) For the same value of angle of incidence, the angles of refraction in three media A, B and C are15°, 25° and 35°, respectively. In which medium, would the velocity of light be minimum? 

(ii) Why does a crack in a glass windowpane appear silvery? 

(iii) The refractive index of diamond is much higher than that of glass. How does a diamond cutter make use of this fact?
                                           OR
What is the apparent position of an object below rectangular block of glass 6 cm thick, if a layer of water 4 cm thick is on the top of the glass?
(Take, ${\mathrm{n}}_{\mathrm{ga}}=1.5 \mathrm{and} {\mathrm{n}}_{\mathrm{wa}}=1.33$).

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{From} \mathrm{Snell}\text{'}\mathrm{s} \mathrm{law}, \mathrm{\mu }=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\frac{\mathrm{c}}{\mathrm{\nu }} \\ \Rightarrow \mathrm{\nu } \propto \sin \mathrm{r}, \mathrm{for} \mathrm{given} \mathrm{value} \mathrm{of} \mathrm{i}. \end{gathered}$$

Smaller the angle of refraction, smaller the velocity of light in medium.

Velocity of light is minimum in medium A, as the angle of refraction is minimum, i.e., ${15}^{\circ }.$

ii) Whenever rays of light travels through glass, they strike the glass-air interface at an angle greater than critical angle of glass. They are totally reflected; hence crack appears silvery. 

(iii) The refractive index of diamond is much higher than that of glass. Due to high refractive index, the critical angle for diamond-air interface is low. The diamond is cut suitably, so that the light entering the diamond from any face suffers multiple total internal reflections at the various surfaces. This gives sparkling effect to the diamonds.

OR

$\mathrm{Here}, \mathrm{\mu }=\frac{\mathrm{real} \mathrm{depth} / \mathrm{thickness} \mathrm{of} \mathrm{object}}{\mathrm{apparent} \mathrm{depth}}$

Now, due to refraction at two different boundaries, the apparent depth of object

$$\begin{gathered} =\frac{\mathrm{thickness} \mathrm{of} \mathrm{glass}}{{\mathrm{\mu }}_{\mathrm{glass}}}+\frac{\mathrm{thickness} \mathrm{of} \mathrm{water}}{{\mathrm{\mu }}_{\mathrm{water}}} \\ =\frac{6}{1.5}+\frac{4}{1.3 } \\ =4+3=7 \mathrm{cm} \end{gathered}$$