A set of narrow vertical slits is located a distance D from a screen. The slits

Question

A set of narrow vertical slits is located a distance D from a screen. The slits are equally spaced and have the same width. The intensity pattern in the fiqure is observed when light from a laser passes through the slits, illuminating them uniformlv. The screen is perpendicular to the direction of the light. Data: Distance to the screen = 2.89 m, wavelength of Tight = 520 nm, Distance between tick marks on the intensity figure = 1.40 cm What is the spacing between the slits? There is a fixed relation between wavelength, slit separation, and angle between the first (or second, third etc) order maximum and the central aximum. The angle can be determined from the distance to the screen and the separation between the first maximum and the central maximum on the screen Submit Answer Incorrect. Tries 1/10 Previous Tries Calculate the width of the slits. There is a fixed relation between wavelength, slitwidth, and angle between the first minimum due to the slit width and the central maximum. There is a difference between the minima due to the slit to slit interference and the minima due to the diffraction due to the slit widths. Compare with the fiqure of the previous problem. The angle can be determined from the distance to the screen and the separation between the first minimum and the centra maximum on the screen. Submit Answer Incorrect. Tries 1/10 Previous Tries If the slit separation is increased by a factor of 3, what would be the distance between the principal peaks on the screen? Submit Answer Tries 0/10

Explanation / Answer

(a)

dsin(theta) = dy/L = lamda

d = lamda*L/y = 520*10^-9 * 2.89/1.4*10^-2

d = 1.07* 10^-4 m

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(b)

a = lamda*L/5y

= 520*10^-9 * 2.89/(5*1.4*10^-2)

= 2.15*10^-5 m

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(c)

y' = lamda*L/d'

= lamda*L/3d

= 520*10^-9*2.89/(3*1.07*10^-4))

= 4.68*10^-3 m

= 0.468 cm

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