Rayleigh Criterion Question

[bananabandana]

Homework Statement

'A scientist wants to take a picture of a distant yellow object using a pinhole camera such that the picture is of maximum sharpness. Let $\lambda$ =wavelength of yellow light, $d$ = diameter of pinhole, $D =$ distance of pinhole to film. Find $d$.

Homework Equations

Rayleigh:
$$ sin(\theta_{R}) = 1.22 \frac{\lambda}{D} $$

The Attempt at a Solution

[/B]
If I can express $\theta_{R}$ in terms of something else this is easy - just apply small angle approximation and rearrange [since know $\theta << 1 rad$].

The answers I've got tell me that:
$$ 2\theta_{R} = \frac{d}{D} $$
But I can't see any geometric way to justify this? Does anyone know where this result comes from? It's just stated as if it's patently obvious, but I've never seen it before...

Thanks!

 

[Charles Link]

You need to consider what a pinhole camera is. Using ray tracing, the light travels in straight lines, so that a finite sized-hole will introduce some blurring. (The pinhole camera basically inverts left and right and up and down. ) The ray (a small light source ) in the object plane puts a small bright spot on the image plane after passing through the pinhole... This tells you that for the sharpest picture, if ray-tracing applies, that you want the (pin)hole as small as possible. Diffraction theory has the opposite effect=the smaller the hole, the greater the angular spread as the wave tries to get through the hole. The optimal case=minimal blurring, occurs when the blurring from diffraction is equal to the ray-trace blurring from the finite sized hole. ...Additional info. : In practice, unless you have very sensitive film or a very bright object, you widen the hole somewhat to get enough light to make an exposure of the film. Normally, cameras use lenses=the pinhole camera is an alternative that because of the low light levels, is rather impractical for photographic use. Perhaps the best application for a pinhole camera is in observing a solar eclipse=you view the image of the sun on the screen after light from the sun passes through the pinhole. If your screen is far enough away from the pinhole, you can get quite a good sized image of the sun as it is being eclipsed. The sun subtends an angle of approximately $\theta=.01 radians$ so that if the screen is one meter from the pinhole, the image of the sun is about 1 centimeter across...