Q. 73

Expert-verifiedFound in: Page 959

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

FIGURE shows two nearly overlapped intensity peaks of the sort you might produce with a diffraction grating . As a practical matter, two peaks can just barely be resolved if their spacing equals the width w of each peak, where is measured at half of the peak’s height. Two peaks closer together than will merge into a single peak. We can use this idea to understand the resolution of a diffraction grating.

a. In the small-angle approximation, the position of the peak of a diffraction grating falls at the same location as the fringe of a double slit: . Suppose two wavelengths differing by pass through a grating at the same time. Find an expression for , the separation of their first-order peaks.

b. We noted that the widths of the bright fringes are proportional to , where is the number of slits in the grating. Let’s hypothesize that the fringe width is Show that this is true for the double-slit pattern. We’ll then assume it to be true as increases.

c. Use your results from parts a and b together with the idea that to find an expression for , the minimum wavelength separation (in first order) for which the diffraction fringes can barely be resolved.

d. Ordinary hydrogen atoms emit red light with a wavelength of In deuterium, which is a “heavy” isotope of hydrogen, the wavelength is What is the minimum number of slits in a diffraction grating that can barely resolve these two wavelengths in the first-order diffraction pattern?

(a) The expression of first order peaks

(b) The double slit pattern is

(c) The diffraction fringes are

(d) The minimum number of slits in a diffraction grating is

We have two wavelengths, and we want to compute the difference between the first brilliant fringe points for each wavelength, where the first brilliant fringe point is

As a result, the following wavelength's placement "(call it "is

Therefore

As given in FIGURE, the peak width is calculated by measuring distances between the point where the intensity is twice of its highest value and the corresponding point on the crest, where the strength of a perfect double slit is supplied by the following formula.

Our goal is to determine the place where the concentration becomes using this equation.

The denominator of both sides yields

To isolate , change the equation.

The breadth of the fringe is double the value of due to the symmetry. So,

However, given we know that , the margin width may be expressed in terms of as tries to follow:

So, for , we showed that the solution is valid.

For minimum changes, we can solve the equation we found in part (a) as

The width of fringed is represented by the equation , and we understand that . so,

To isolate , substitute and modify the equation.

Now may rephrase the number of slits but using the calculation we found in part (c).

We need to divide the smaller wavelength by to get the smallest number of incisions that can barely resolve the two supplied wavelengths. Therefore

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