Diffraction from a single slit, with Lens

[JosephK]

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.

 

[Simon Bridge]

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.

 

[elette]

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 realize 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 that's right! Good luck all!

 

[Simon Bridge]

@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?

 

[Nickie]

But, assuming it is, the solution approach is correct.

To find the width of the slit, we can use the equation a/2 sin theta = lambda / 2, where a is the width of the slit, theta is the angle of diffraction, and lambda is the wavelength of the light. In this case, we know that the distance from the central maximum to the first minimum is 10.2 mm, which corresponds to the angle of diffraction. We also know that the wavelength of the light is 546 nm.

Using the tangent function, we can find the value of theta by taking the inverse tangent of 10.2 mm / 600 mm. This gives us a value of approximately 0.0095 radians. Plugging this value into the equation, we can solve for the width of the slit, a.

a/2 sin 0.0095 = 546 nm / 2

a = (2 * 546 nm * sin 0.0095) / 2

a = 10.3 nm

Therefore, the width of the slit is approximately 10.3 nm. This small width allows for the diffraction of light, creating the pattern seen in the focal plane of the lens.