Light consisting of two wavelengths, a = 5.1 x 10 -7 meter and b = 5.7 x 10 -7 m

Question

Light consisting of two wavelengths, a = 5.1 x 10-7 meter and b = 5.7 x 10-7 meter, is incident normally on a barrier with two slits separated by a distance d. The intensity distribution is measured along a plane that is a distance L = 0.80 meter from the slits as shown above. The movable detector contains a photoelectric cell whose position y is measured from the central maximum. The first order maximum for the longer wavelength b occurs at y = 1.4 x 10-2 meter.

Photoelectric Detector Light Plane of Measurements L. Note: Figure not drawn to scale Light consisting of two wavelengths, Ma 5.1 x 10 7 meter and Ab 5.7 x 10 7 meter, is incident normally on a barrier with two slits separated by a distance d. The intensity distribution is measured along a plane that is a distance L 0,80 meter from the slits as shown above. The movable detector contains a photoelectric ce whose position y is measured from the central maximum. The first order maximum for the longer wavelength Ab occurs at y 1.4 x 1. meter. a. Determine the slit separation d. b. At what position ya does the first order maximum occur for the shorter wavelength Aa?

Explanation / Answer

As we know that

n = yd/L, where

is the wavelength of the light,
d is the separation of the slits,
n is the order of maxima observed (central maximum is n=0),
y is the distance between the bands of light and the central maximum (also called fringe distance), and
L is the distance from the slits to the screen.

So,

d = n*L / y

d = 1*5.7*10^-7*0.8 / 1.4*10^-2

d = 3.257143*10^-5 m

b)

Again we know that

n = yd/L

1*5.1*10^-7 = y *3.257143*10^-5 /0.8

y = 1.252631524*10^-2 m

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