- #1
eep
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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
[tex]\frac{h^2n-t}{h}[/tex]
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:
[tex]D_r\frac{n_1}{n_2} = D_a[/tex]
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.
"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
[tex]\frac{h^2n-t}{h}[/tex]
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:
[tex]D_r\frac{n_1}{n_2} = D_a[/tex]
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.