Solving Geometric Optics Problem: Raising Height of Camera

[eep]

Here's a question that everyone in my class that I talked to couldn't find an answer to.

"Suppose that you focus a camera direclty down on a printed letter on this page. The letter is then covered with a 1.00mm thick microscope slide (n = 1.55). How high must the camera be raised in order to keep the letter in focus?"

I derived a formula for the new focusing distance that depends on the original height that it was focused at by comparing two triangles with the same angle, one with the base being half that of a letter and the other with the base of half a letter minus the width that the light ray is shifted from passing through the glass.

The formula is

\frac{h^2n-t}{h}

where h is the original focus distance, n is the index of refraction of the glass, and t is the thickness of the glass.

The solution that was posted gives a finite distance and is solved using a formula derived from a previous problem to prove that an object in water looks 3/4 as deep as it really is. The formula is:

D_r\frac{n_1}{n_2} = D_a

where Da is the apparent distance and Dr is the real distance to an object.

Basically I'm wondering how someone else would approach the problem, because the solution given to us didn't really fit well with me.

 

[lightgrav]

If I understand your description (and I might not)
you've found tha new camera height that would
make each letter the same (angular) size at the lens,
to compensate for Snell's deflection of the top ray.

This would give both photos the same field-of-view
(a much more useful and interesting calculation),
but does not address whether the camera would
need to be "re-focused" because of the glass plate.

I would've treated the lens as large diameter, close,
with 2 rays converging on the center of the view,
after being deflected by Snell's Law (small angle);
that's how I'd do the "apparent depth in water" too.

 

[eep]

I would've treated the lens as large diameter, close,
with 2 rays converging on the center of the view,
after being deflected by Snell's Law (small angle);
that's how I'd do the "apparent depth in water" too.

Could you explain this a little more? We were assuming that the letter must appear to be the same size before and after the glass slide is placed.

 

[lightgrav]

But the question only says "keep it in focus".

Look with 2 eyes, stereo vision, same h from letter.
rays form 2 similar right triangles, one to each eye.
With a fixed eye spacing, find theta at the eyes.
(here, the eyes are opposite edges of the lens)

Keep theta at the eyes the same, with the slide there.
Snell says sin(theta) = n_glass sin(theta_glass),
so theta_glass approx theta/n_glass.
The page has to move farther away or else
the rays in glass (closer to Normal) don't form triangles.

Isn't that how your book derived the water depth?

 

[eep]

our book didn't derive anything to do with that. the way it's presented in the homework solutions that were posted (worked out by hand by a grad student) use one eye placed at the center.

for the water case, one diagram was drawn with both air and water and the light rays (with bending) coming from the corner of the object and making one triangle from the edge of the object to the water surface, and another from the water surface to a central eye above the air. the distance from the object to the water surface is called the actual distance. then, we imagine the water dissapears and trace the rays from the eye, so that one can form a new triangle with the same width as the triangle below the water surface, but with the angle in the air instead of the angle in the water. using some algebraic trickery and the assumption that the cosine of either angle is 1 (small angles) you end up with the formula in my first post. i think i understand it much better now though.