Light with a wavelength of 600 nm passes through a double slit and an interference pattern is observed on a screen 2.00 m from the slits. The first-order bright fringe is at 4.00 mm from the center of the central bright fringe. For what wavelength will the first order dark fringe be observed at the same point on the screen?

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The figure is below attached. The light is coming from the left on the figure and arrives at the double slit. Each slit will act as an independent source of light. If the wave front of the incoming wave is parallel to the slits, the light radiated from the two slits will be in phase. Thus at a point on the screen on the right the difference of phase between the two waves radiated by the two slits will be given by their geometrical path difference (see the figure):

`delta =d*sin(theta)`

Since from geometrical considerations in the figure we have

`tan(theta) = L/D` and for small angles one can approximate

`tan(theta) =sin(theta) (=theta)`

one can write the path difference as

`delta = d*L/D`

The condition for obtaining a `k` order bright fringe (thus maximum of light interference) is `delta =k*lambda` while the condition for `k` order dark fringe (destructive interference of light) is `delta =(2*k+1)*lambda/2`

From the text one has for `k =1` , `L =4 mm` and `lambda =600 nm` for bright fringe.

`d*L/D = 1*lambda`

`d*0.004/2 =600*10^-9` , which means

`d =0.0003 m =0.3 mm`

For the first order dark fringe one has the data `k=1` , `L =4 mm` , `d=0.3mm`

Thus the corresponding wavelength is

`d*L/D =(2*1+1)*lambda_1/2`

`lambda_1 =(2/3)*d*L/D =2/3*0.0003*0.004/2 =4*10^-7 m =400 nm`

**The wavelength of light for which the first order dark fringe is at the same position as the first order bright fringe is 400 nm.**

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