Sunlight is used in a double-slit interference experiment. The fourth-order maxi

Question

Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 480 nm occurs at an angle of theta = 90 degree. Thus, it is on the verge of being eliminated from the pattern because theta cannot exceed 90 degree 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? (a) Number Units (b) (c) Number Units

Explanation / Answer

for 4th order

from d sin theta = m L

d = 4 * 480 nm/sin 90

d = 1920 nm or 1.92 um

so

now use d sin theta = mL

here wavelength L = ?

m = 3

theta = 90 deg

L = 1.92 *10^-6 * sin 90/3

L = 640 nm

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b. as m increases, d decrases

to decrease m, d must increases

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d2-d1 = least separation = 1920 nm - 640 nm

d2-d1 = 1280 nm o r 1.28 um

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