(a) A plane wavefront $AC$ is incident on the plane of separation XY of two media of refractive indices $\mu_1$ and $\mu_2 (\mu_2>\mu_1)$ making an angle $i$. This is known as angle of incidence. When the wavefront touches the point $A$, the point becomes a source of secondary wavelets. Thus, when the whole waveform passes through the $X Y$ plane, each point of AF becomes the source of secondary wavelets. When point $C$ of the wavefront in medium 1 traverses CF distance by that time $(t)$ the wavelet from point $A$ traverses $AD$ distance. If $v_1$ and $v_2$ are the speeds of light in medium 1 and 2 respectively, then $AD = v_2 t$ and $CF = v_1 t$. Refracted wavefront DF which is obtained by drawing a tangent to the arc having radius $v_2 t$ and centre $A$. The angle made by the tangent with the plane $XY$ is $r$. This is known as angle of refraction. The perpendiculars drawn on wavefront $AC$ are the incident rays. The perpendiculars drawn on wavefront $DF$ are the refracted rays. $AN$ and TF are the perpendiculars drawn on $XY$, the plane of separation of the two media. $\angle CAF = \angle i = 90^{\circ} - \angle NAC$ $= 90^{\circ} - (90^{\circ} - \angle SAN)$ $\therefore \angle SAN = \angle i$ $\text{ Similarly, } \angle QFT = \angle r$ $\text{ In } \triangle ACF,$ $\sin i = \frac{CF}{AF} = \frac{v_1 t}{AF}$ In $\triangle ADF$, $\sin r = \frac{AD}{AF} = \frac{v_2 t}{AF}$ $\therefore \frac{\sin i}{\sin r} = \frac{c_1 t}{c_2 t} = \frac{c_1}{c_2} = {}_1\mu_2$ This is Snell's law. OR (b) A plane wavefront $AC$ is incident on a plane reflector $XY$ making an angle $i$. This is known as angle of incidence. Each and every point of the wavefront when touches the reflector becomes a source of secondary wavelets. When the wavefront touches the point $A$, the point becomes a source of secondary wavelets. Thus, when the whole waveform touches the XY plane, each point of AF becomes the source of secondary wavelets. When point $C$ of the wavefront in medium 1 traverses CF distance by that time $(t)$ the wavelet from point $A$ traverses $AD$ distance. If $v_1$ is the speeds of light in medium then $AD = v_1 t$ and $CF = v_1 t$. Reflected wavefront DF which is obtained by drawing a tangent to the arc having radius $v_1 t$ and centre $A$. The angle made by the tangent with the plane $XY$ is $r$. This is known as angle of refraction. The perpendiculars drawn on wavefront $AC$ are the incident rays. The perpendiculars drawn on wavefront DF are the reflected rays. $AN$ and TF are the perpendiculars drawn on $XY$, the plane reflector.$\angle CAF = \angle i = 90^{\circ} - \angle NAC$$= 90^{\circ} - (90^{\circ} - \angle SAN)$$\therefore \angle SAN = \angle i$$\text{ Similarly, } \angle QFT = \angle r$In $\triangle ACF$ and $\triangle AFD$$\angle ACF = \angle ADF = 90^{\circ}$$CF = AD$$AF$ is the common sideSo, the triangles are congruent.$\therefore \angle CAF = \angle AFD$$\therefore i = \angle r$This is law of reflection.