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Consider red light with a wavelength of 692 nm in air that is incident onto two slits such that it gives diffraction fringes that are 1.00 mm apart on a screen at a distance L from the slits. If the screen is moved back by an additional 4.00 cm, the fringes become 1.20 mm apart. What is the separation between the slits?

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Answer:

The separation between slits is:


d=0.1384 \, mm

Step-by-step explanation:

The distance between adjacent maxima is given by the following expression:


\Delta y=(\lambda L)/(d)

For the distance between maxima being 1mm (let's solve all this problem using milimeters) we have the following:


1=(L\lambda)/(d) (1)

For the second condition in which the screen is moved 40 mm away we have the following:


1.20=((L+40)\lambda)/(d) (2)

Using equation 1 we can write:


L=(L)/(d)

Plugging the value of L in equation 2 brings us to


1.20=(\lambda\left((d)/(\lamda)+40\right))/(d)


\implies 1.20=(d+40\lambda)/(d)


\implies 1.20d=d+40\lambda


\implies d=(40\lambda)/(0.20)=0.1384\, mm

We could also as a bonus find L which is just


L=(d)/(\lambda)

I'll leave that up to you! Good luck!

Consider red light with a wavelength of 692 nm in air that is incident onto two slits-example-1
User Amin Arghavani
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