Diffraction from a single slit, with Lens

  • Thread starter JosephK
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Homework Statement


Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit covering a lens with a focal length of 60.0 cm. In the focal plane of the lens the distance from the central maximum to the first minimum is 10.2 mm. What is the width of the slit


Homework Equations



a/2 sin theta = lambda / 2


The Attempt at a Solution



Tangent theta is opposite over adjacent... The adjacent, I think, is the focal length, 60.0 cm. The opposite is clearly 10.2 mm. So, tan theta = 10.2 mm / 600 mm. Then to get a, the width of the slit, I use the above equation.

I am not familiar with lenses. So, I am not sure if the "adjacent" is correct.
 

Answers and Replies

  • #2
Simon Bridge
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Hmmm ... without the lens, dsinθ=nλ and Ltanθ=Δx: L is the distance to the screen and d is the separation of the slits.

In the limit L >>d, what happens to these equations?

The effect of the slits is to produce maxima "rays" that are diverging: they fan out.
The effect of the lens is, presumably, to converge the rays - so they fan out less.
Your calculation would be for the same situation without the lens.

So you need to be able to relate the angle of the ray incident to the lens with the angle of the emerging ray.
 
  • #3
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I have just tried this question and I believe I have the answer - not sure if you have by now, but for any others out there....

The trick is to realise how to derive the formula for the fringe spacing of minima produced by the single slit.
y = m\lambda L / a
In that derivation, L was used as the distance to the screen- With a lens: L must be equivalent to the focal length, f. So that y = fringe spacing = (m\lambda f)/a . Where a is equal to the slit width, \lambda : wavelength of light.
Hope thats right! Good luck all!
 
  • #4
Simon Bridge
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@elette:
Welcome to PF;
In this problem the slit is right on the lens and the screen is in the focal plane - so L=f ... fine.
Thinking of the derivation of the fringe-spacing equation is important - however:
In that derivation, L was used as the distance to the screen- With a lens: L must be equivalent to the focal length, f. So that y = fringe spacing = (m\lambda f)/a . Where a is equal to the slit width, \lambda : wavelength of light.
... $$\Delta y = \frac{\lambda f}{a}$$ ... does not appear to account for the refraction of the light through the lens.
Does the lens have no effect at all?
 

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