Chapter 28: Physical Optics: Interference and Diffraction Selected Solutions

Selected Solutions

7. The path differences equal to one and two wavelengths correspond to the first and second lowest frequencies, respectively. The path difference is given by

[distance from the furthest speaker] - [distance from the nearest speaker]. Therefore,

Lowest frequency is for path difference = l:

Next lowest frequency is for path difference = 2l:

13. (a) For a second order (m = 2) maximum the condition is: . Thus,

.

(b) Since the wavelength is directly proportional to the separation, the wavelength will increase if the separation is increased.

(c) For this larger slit separation we get

which is indeed larger as stated in part (b).

31. (a) Since , there is no phase change at the coating-lens boundary. The condition for destructive interference is

The first few results give

The only wavelength that is within the visible range is l = 503.2 nm which is the one that will be absent.

(b) Since , there is a phase change at the coating-lens boundary. Adding 1/2 to the left side of the equation for destructive interference will account for the reflection at the coating-lens interface.

l = 670.9 nm and 402.6 nm will be absent.

39. The last dark fringes theoretically occur where q approaches

So, there are 14 dark fringes produced on either side of the central maximum.

75. The smallest diameter corresponds to the minimum angle of resolution. This angle of resolution is subtended by the radius of the pupil. Therefore, diameter = 2y = 2Ltanq_min. The angle can be determined by

The diameter, then, is given by

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