Two lasers are shining on a double slit, with slit separation d . Laser 1 has a

Question

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 5.90 m away from the slits.

Part A

Which laser has its first maximum closer to the central maximum?

Part B

What is the distance ymaxmax between the first maxima (on the same side of the central maximum) of the two patterns?

Express your answer in meters.

Part C

What is the distance ymaxmin between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Express your answer in meters.

Explanation / Answer

a) The first maxima comes when the path difference between the two slits is equal to one full wavelength.

   So laser 1 has its first maximum closer to the central maximum.

b) The first maxima corresponds to constructive interference with m = 1

Now for laser 1,   d*sin(theta) = m*lambda

                              d*sin(theta) = 1*d/20

                               sin(theta) = 1/20

y = distance on the screen from central maximum = R*sin(theta)

                     y1 = 5.9/20 = 0.295 m

Similarly for laser 2,   d*sin(theta) = m*lambda

                              d*sin(theta) = 1*d/15

                               sin(theta) = 1/15

y = distance on the screen from central maximum = R*sin(theta)

                     y2 = 5.9/15 = 0.393 m

Difference = y2 – y1 = 9.8*10^-2 m

c) The second maxima corresponds to constructive interference with m = 2

Now for laser 1,   d*sin(theta) = m*lambda

                              d*sin(theta) = 2*d/20

                               sin(theta) = 1/10

y = distance on the screen from central maximum = R*sin(theta)

                     y1 = 5.9/10 = 0.59 m

Similarly for laser 2,   d*sin(theta) = (m+0.5)*lambda

                              d*sin(theta) = 2.5*d/15

                               sin(theta) = 2.5/15 = 0.167

y = distance on the screen from central maximum = R*sin(theta)

                     y2 = 5.9*0.167 = 0.9853 m

Difference = y2 – y1 = 0.3953 m

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