FULL SCREEN PRINTERVERSION ACK NExT Chapter 35, Problem 026 Sunlight is used in

Question

FULL SCREEN PRINTERVERSION ACK NExT Chapter 35, Problem 026 Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 500 nm occurs at an angle of 0 90°. Thus, it is on the verge of being eliminated from the pattern because 6 cannot exceed 90 in Eq. 35-14. (a) What least wavelength in the visible range (400 nm to 700 nm) are not 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 (a) Number (b) units (c) Number Click if you would like to Show Work for this question Question Attempts: 0 of 5 used Copyright 2000-2017 by John Wiley & Sons, Inc. or related companies. All rights

Explanation / Answer

d*sintheta = m*lambda

d = m*ambda/sin90

d = 4 * 500 = 2000 nm

for 3rd order

lambda = d*sin90/m = 666.67 nm

any wavelength of 666.67 not seen so 666.67 nm < theta < 700 nm are not present

part b )

d should be decreased

part d )

dnew *sintheta = m*lambda

in this case lambda = 400 nm , m =4

dnew = 1600 nm

difference = 2000 - 1600 = 400 nm

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