The X–Y plane be taken as the boundary between two transparent media M1 and M2.M1 in Z 0 has a refractive index of \(\sqrt2\) and M2 with \(Z<0\) has a refractive index of \(\sqrt3\). A ray of light travelling in M1 along the direction given by the vector \(\mathbf{\overrightarrow{P}} = 4\sqrt{3}\mathbf{\hat{i}} - 3\sqrt{3}\mathbf{\hat{j}} - 5\hat{k}\), is incident on the plane of separation. The value of difference between the angle of incident in M1 and the angle of refraction in M2 will be ______ degree.
Normal will be \(−\hat{k}\) so \(\cos i = \frac{\mathbf{P} \cdot \hat{n}}{\left|\mathbf{P}\right| \cdot \left|\hat{n}\right|}\) \(\frac{5}{10} = \frac{1}{2}\) \(⇒ i = 60°\) Using snells law \(\sqrt{2} \sin 60^\circ = \sqrt{3} \sin r\)
\(\frac{\sqrt{3}}{\sqrt{2}} = 3 \sin r\) \(⇒ r = 45°\) So, \(i – r = 60°-45°\) \(= 15°\) So, the answer is 15°.
Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume that S1 and S2 are two coherent sources derived from S.
The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S1 and S2. It is considered that d is the separation between both the slits. The S1 is opened, S2 is closed and the screen opposite to the S1 is closed, but the screen opposite to S2 is illuminating.
Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.
