Question

A lens rest on a flat surface as shown in figure (0.4).

Determine the radius of the nth bright spot.

Figure 0.4: Problem 6

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Answer #1

A R1R2 PL.

AB is a monochromatic ray of light, which falls on the system. A part is reflected at B (glass-air boundary) which goes out in the form of ray R1 without any phase reversal. The other part is refracted along BC. At point C it is again reflected and goes out in the form of ray R2 with a phase reversal of . The reflected rays R1 and R2 are in a position to produce interference fringes as they have been derived from the same ray AB and hence fulfill the condition of interference. As the rings are observed in the reflected light, the path difference between them is (2µt cos r + lambda/2). For air film µ = 1 and for normal incidence r = 0. Hence in this case, path difference is (2 t + lambda/2). At the point of contact t = 0, and the path difference is lambda/2, which is the condition of minimum intensity. Thus the central spot is dark. For nth maximum, we have

2 t + lambda/2 = n * lambda

This expression shows that a maximum of a particular order will occur for a constant value of t. In this system, ‘t’ remains constant along a circle Thus the maximum is in the form of a circle. For different value of ‘t’, different maxima will occur. Hence we get a number of concentric bright circular rings. In a similar way, this can be shown that minima are also in the circular form.

Theory :

2. O L 2. Siurau 2. a R 2. Can- àR Spot

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