Read, Study&Practice; Assignment Open Assignment Resources Chapter 35, Problem 0

Question

Read, Study&Practice; Assignment Open Assignment Resources Chapter 35, Problem 026 Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 490 nm occurs at an angle of 0 90°. Thus, it is on the verge of being eliminated from the pattern because present in the third-order maxima? To eliminate all of the visible light in the fourth-order maximum (b) should the slit separation be increased or decreased and (c) what least change in separation is needed? units Number Pro tem (b) c) Number Click if you would like to Show Work for this question: Question Attempts o of s used SAve on Area SARMITANSAERI 2000 2017 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Explanation / Answer

(a)

we have sin = n/d for maximum, where d is separation between the slits.

So, here,  sin900 = 4(490nm)/d

or, d = 1960nm

Now, for third order maxima, n = 3

So, let wavelength   is not visible for which   = 900, so we get

sin900 = 3()/d

or, = d/3 = 1960nm/3 = 653.33 nm

So least wavelength not present in third order maxima is 653.33 nm

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(b)

for a given wavelength, the distance of maxima y from the center of the bright fringes increases with decreasing separation between slits. To eliminate a certain wavelength we have to make y large engough for that wavelgnth so that the maxima is out of the screen

So, we should decrease the slit separation.

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(c)

we know, y = nD/d

Initially y = 4(D/d1)(490 nm) --------- (1)

to eliminate all the visible light we sould have y same as above for = 400 nm

so, y = 4(D/d2)(400 nm) -------------- (2)

from (1) and (2), we get,

400nm/d2 = 490nm/d1

or, d2 = (400nm/490nm)d1 = (400nm/490nm)(1960nm)

or, d2 = 1600nm

So decrease in slit separation needed is, d1 - d2 = 1960nm - 1600nm = 360nm

The change or decrease in slit separation is 360 nm

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