Subject: Applied Physics 2

Topic: Interference And Diffraction Of Light

Difficulty: Medium

i)The angle of wedge film enclosed between glass plate and lower surface of lens is very small and hence can be neglected. Since the angle is small, thickness of air film trapped is less .so higher order can be neglected.

ii) It is seen that the diameter of dark rings is given by $D_n^2$ = 4nλR .

That is diameter of dark ring is proportional to square root of the natural numbers .therefore, the diameter of the ring does not increase in the same proportion as the order of the ring.For example , if n increases as 1,2,3,4......, the diameters are $D_1$ =2 $\sqrt {λR}$, $D_2$ =2 (1.4) $\sqrt{λR}$ , $D_3$ =2 (1.7) $\sqrt{λR}$,$D_4$ =2 (2) $\sqrt{λR}$ and so on ..... Therefore the rings get closer and closer as n increases

       iii) In newtons ring, the optical path difference in reflected light is given by
                      Δ =2μtcos(r+ϴ)± λ/2 .

At the point of contact of lens and the plate thickness is zero, so optical path difference becomes only λ/2 which is condition for destructive interference. Therefore, the centre is dark.

iv)When a planoconvex lens of large radius of curvature is placed on a flat glass plate, a thin film of air is trapped between lens and plate. The thickness of the film is such that thickness is zero at the point of contact of lens and plate and gradually increases outwards. The locus of all the points having same thickness falls on the circle.so the air film has a circular symmetry. When this is illuminated alternate dark and bright concentric rings are observed. So the fringes are circular.