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Laser light of wavelength 500.0 nm illuminates two identical slits, producing an interference pattern on a screen 90.0 cm from the slits. The bright bands are 1.00 cm apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

Short Answer

Expert verified

The width and separation of the two slits are 15.0μmand45.0μm.

Step by step solution

01

Formulas used to solve the question

Angle of the minimum fringes due to the single-slit diffraction:

asinθ=mmin (1)

Angle of the bright fringes due to the double-slit interference:

dsinθ=mmax (2)

02

Determine the width of the two slits

Given: λ=500.0nm=500.0*10-9m

R=90.0cm=90.0*10-2mΔy=1.0cm=1.0*10-2m

Angle of the minimum fringes due to the single-slit diffraction:

asinθ=mmin

Angle of the bright fringes due to the double-slit interference:

dsinθ=mmax

Divide the equation (2) by (1),

da=mmaxmmin

The third bright is for mmaxand since it is the first missing bright fringe, so it meets the first dark fringe of the single slit diffraction which is for mmin

Therefore,

d = 3a

(3)

Now, the distance from the central maxima to the third minimum is given by,

y3=3Δy

Now, find the angle since it is the same angle for both equations (1) and (2) and plug the given,

tanθ3=y3R=3ΔyRθ=tan-1(3ΔyR)=tan-1(3*1.0*10-290*10-2)=1.91

Now, solve (1) forand plug the given,

a=λsinθ=500*10-9sin1.91°15.0μm

03

Determine the separation of the two slits

Solve (2) for d and plug the given,

d=3λsinθ=3*500*10-9sin1.91°45.0μm

Thus, the width and separation of the two slits are 15.0μmand45.0μm.

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