# Double slits and convex lens interference

• fedecolo
The Attempt at a SolutionI don't know what to do because I don't know what is the effect of the lens on the maximums and minimums.

## Homework Statement

Two slits (of width ##a=39 \mu m##) are lighted up with a monocromatic wave of ##\lambda=632,8 nm##. The distance between slits and the screen id ##D=4 m##. The distance between the slits id ##d=195 \mu m##.
In front of the slits there are a convergent lens with focal length ##f=60 cm##.

Find the width of central maximum of interference and diffraction.

## The Attempt at a Solution

I don't know what to do because I don't know what is the effect of the lens on the maximums and minimums.

Without lens the width of central maximum (of interference) is the distance between the first two minimum, so $$d\sin\theta=\frac{1}{2}\lambda \longrightarrow d\frac{y_{min}}{D}=\frac{1}{2}\lambda \longrightarrow \Delta y=2y=\frac{\lambda D}{d}$$

and for the diffraction it would be $$a\sin\theta=\lambda \longrightarrow a\frac{y_{min}}{D}=\lambda \longrightarrow \Delta y=2y=2\frac{\lambda D}{a}$$.

Can you give me a hint?

It's not mentioned whether the lens is between the source and the slits, or between the slits and the screen.
In the former case it shold be apparent that there is no effect since both slits act as point sources no matter what the incident light.
In the latter case, if 2 parallel rays impinge on a thin lens, is there any difference in path length between them after they exit the lens? So what if any effect is there between having the lens or not?

(Extra credit: the lens has a beneficial effect if d is not << D. What is it?)

rude man said:
It's not mentioned whether the lens is between the source and the slits, or between the slits and the screen.
In the former case it shold be apparent that there is no effect since both slits act as point sources no matter what the incident light.
In the latter case, if 2 parallel rays impinge on a thin lens, is there any difference in path length between them after they exit the lens? So what if any effect is there between having the lens or not?

(Extra credit: the lens has a beneficial effect if d is not << D. What is it?)
We are in the latter case.
And no, the difference of path remains the same before the rays impinge on the lens and after they exit. But this fact doesn't give me no information about the position of the maximums/minimums of interference.

fedecolo said:
We are in the latter case.
And no, the difference of path remains the same before the rays impinge on the lens and after they exit. But this fact doesn't give me no information about the position of the maximums/minimums of interference.
Since there is no path difference between rays, doesn't that suggest that they'll focus on the same point as they would have had there been no lens?

Cf. my next post.

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From my Resnick & Halliday

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rude man said:
Since there is no path difference between rays, doesn't that suggest that they'll focus on the same point as they would have had there been no lens?

Cf. my next post.
I thought the same thing, but without the lens I would have a width of ##\Delta y_{interference}=1,3 cm## and with the lens the answer should be ##\Delta y_{interference}=\Delta y_{diffraction}=1,95 mm##.

By seeing the picture, I think that the rays focus on the same points (of the case without lens) iff ##D=f## (distance between slits and screen is ##f##), but now I have to consider that there is another distance between my "new" screen (at distance ##f##) and the "old" screen at distance ##D##.

fedecolo said:
I thought the same thing, but without the lens I would have a width of ##\Delta y_{interference}=1,3 cm## and with the lens the answer should be ##\Delta y_{interference}=\Delta y_{diffraction}=1,95 mm##.

By seeing the picture, I think that the rays focus on the same points (of the case without lens) iff ##D=f## (distance between slits and screen is ##f##), but now I have to consider that there is another distance between my "new" screen (at distance ##f##) and the "old" screen at distance ##D##.
You're entirely right, and I mis-stated - jumped the gun.

For the lens-less case the expression is d sinθ = (m+1/2)λ and sinθ = y/D
For the lens case the expression is still d sinθ = (m+1/2) but with sinθ = y/f. Can you make that out in the figure I sent you? I know it's not very legible. The essence is that y/f = tanθ instead of y/D = tanθ. And in both cases we approximate tanθ = sinθ since θ is assumed small.

In other words, with the lens the spacings on the screen is proportional to f instead of D. The lens case does remove the requirement (assumption) that d << D however so it's better in that sense, and therefore usually used.

I'll look at the diffraction case later today.

It would seem to unnecessarily complicate things by putting a convergent lens in the system if the screen is not placed in the focal plane of the lens. Meanwhile, the screen is far enough away at ## D=4 \, m ## that it is in the far field, (without any lens). The convergent lens with ## f=60 \, cm## , IMO, does not belong there and with a screen now out of the focal plane, the interference pattern will likely be a scrambled blur with little to learn from this problem.