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Chapter 36: Q3Q (page 1107)

For three experiments, Fig.36-31 gives the parameter of Eq. 36-20 versus angle in two-slit interference using light of wavelength 500 nm. The slit separations in the three experiments differ. Rank the experiments according to (a) the slit separations and (b) the total number of two slit interference maxima in the pattern, greatest first.

Short Answer

Expert verified
  1. The rank will be A, B and C
  2. The rank will be C, B and A.

Step by step solution

01

The given data

Given for three experiments, the parameter尾 of versus angle胃 in two-slit interference using light of wavelength 500 nm.

The slit separations in the three experiments differ.

02

Concept and Formula used

In a double split experiment for diffraction,

尾=蟺d位sin胃

dis slit separation , and位 is wavelength.

03

(a) Rank according to slit separation

Now, from the equation

尾=蟺d位sin胃

It is clear that graph will be greater for larger values of d.

So Rank according to slit width (greatest first) is A, B and C.

04

(b) Rank according to total diffraction maxima

Since the wavelengths are same, the smaller slit will produce more maxima.

So, the rank will be C, B and A.

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Most popular questions from this chapter

What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 胃=45掳?

At night many people see rings (called entoptic halos) surrounding bright outdoor lamps in otherwise dark surroundings. The rings are the first of the side maxima in diffraction patterns produced by structures that are thought to be within the cornea (or possible the lens) of the observer鈥檚 eye. (The central maxima of such patterns overlap the lamp.) (a) Would a particular ring become smaller or larger if the lamp were switched from blue to red light? (b) If a lamp emits white light, is blue or red on the outside edge of the ring?

If Superman really had x-ray vision at 0.10nm wavelength and a 4.0mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.0cm to do this?

Floaters. The floaters you see when viewing a bright, featureless background are diffraction patterns of defects in the vitreous humor that fills most of your eye. Sighting through a pinhole sharpens the diffraction pattern. If you also view a small circular dot, you can approximate the defect鈥檚 size. Assume that the defect diffracts light as a circular aperture does. Adjust the dot鈥檚 distance L from your eye (or eye lens) until the dot and the circle of the first minimum in the diffraction pattern appear to have the same size in your view. That is, until they have the same diameter D'on the retina at distance L'=2cmfrom the front of the eye, as suggested in Fig. 36-42a, where the angles on the two sides of the eye lens are equal. Assume that the wavelength of visible light is 位=550nm. If the dot has diameter D=2.0mmand is distance L=45.0cmfrom the eye and the defect is x=6.0mm in front of the retina (Fig. 36-42b), what is the diameter of the defect?

Consider a two-dimensional square crystal structure, such as one side of the structure shown in Fig. 36-28a.The largest interplanar spacing of reflecting planes is the unit cell size ao. Calculate and sketch the (a) second largest, (b) third largest, (c) fourth largest, (d) fifth largest, and (e) sixth largest interplanar spacing. (f) Show that your results in (a) through (e) are consistent with the general formula

d=aoh2+k2

where h and k are relatively prime integers (they have no common factor other than unity).
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