# distance of camera lins from mirror.

• Mar 3rd 2008, 12:51 PM
rcmango
distance of camera lins from mirror.
You are trying to photograph a bird sitting on a tree branch, but a tall hedge is blocking your view. However, as the drawing shows, a plane mirror reflects light from the bird into your camera. If x = 3.0 m and y = 4.2 m in the drawing, for what distance must you set the focus of the camera lens in order to snap a sharp picture of the bird's image?
m

heres what it looks like: http://img125.imageshack.us/img125/1508/p2506altfk2.gif

to me it seems I could use trig to find the distance, correct?

its just, none of the distances that are given, I couldn't manipulate them to work for the problem.

• Mar 3rd 2008, 01:58 PM
CaptainBlack
Quote:

Originally Posted by rcmango
You are trying to photograph a bird sitting on a tree branch, but a tall hedge is blocking your view. However, as the drawing shows, a plane mirror reflects light from the bird into your camera. If x = 3.0 m and y = 4.2 m in the drawing, for what distance must you set the focus of the camera lens in order to snap a sharp picture of the bird's image?
m

heres what it looks like: http://img125.imageshack.us/img125/1508/p2506altfk2.gif

to me it seems I could use trig to find the distance, correct?

its just, none of the distances that are given, I couldn't manipulate them to work for the problem.

In the diagram add the image, which is $\displaystyle 2.1$m on the other side of the mirror
to the real bird. That gives you a right triangle with sides $\displaystyle x+2.1$ and $\displaystyle y$, the
hypotenuse is the distance to set the focus and so is:

$\displaystyle d=\sqrt{y^2+(x+2.1)^2}$

RonL
• Mar 3rd 2008, 02:14 PM
ath3na
homework guideance...
I looked at using trig and it looked uglybugly, good old fashioned plane geometry seemed to do the trick with minimal fuss...

I liked CaptainBlack's approach, very neat(Clapping)...but I wrote the following before I saw his reply(Doh)...and since it doesn't rely on knowing about imaginary images behind mirrors and uses different maths I thought I might still post it for it's two-pence worth...may help someone else...

Angle of incidence = angle of reflection, from this you have two similar right-angle triangles, one with sides x and part of mirror, the other with side 2.1m and other part of mirror. The two sides along the mirror add up to y (4.2m), the distance you are looking for is the sum of the hypotenuses of these two similar right-angle triangles.

Remember that for similar triangles all lengths of the similar sides are proportional by the same constant; so find the proportionality for the sides perpendicular to the mirror (this is easy because both are given), use this constant to find the lengths of the sides on the mirror (this is now easy, just substitute either side into the formula that says they both add to y=4.2m). You should now have one of the lengths of the hypotenuse, you get the other using your similar triangles scale factor, then you add them.

Having calculated the sharpest focus the bird will inevitably have moved! See if you can now do the same thing but leave x and y undetermined so that you get a formula which works for any value of x and y you plug in (maybe even for arbitrary distance of bird from mirror)....

Alternatively, give up maths and buy and auto-focus camera...but then how would you get a grasp on the most powerful language mankind has ever known???(Cool)
• Mar 3rd 2008, 02:44 PM
Soroban
Hello, rcmango!

Edit: Captain Black's is brilliant!
I totally forgot about that "reflection trick' . . . (*kick self*)

Quote:

You are trying to photograph a bird sitting on a tree branch,
but a tall hedge is blocking your view.
However, as the drawing shows, a plane mirror reflects light from the bird into your camera.
If x = 3.0 m and y = 4.2 m, for what distance must you set the focus of the camera lens
in order to snap a sharp picture of the bird's image?

We can solve this without trig . . .
Code:

          E  2.1  D       + - * - - - * . . . :    *    | . . . :      *  | a . . . :        * | . . . :          * C . 4.2 :        * | . . . :      *  | . . . :    *    | 4.2-a . . . :  *      | . . . : *        | . . . * - - - - - * . . . A    3.0    B

Note that: .$\displaystyle \Delta CDE \sim \Delta CBA$

. . Hence: .$\displaystyle \frac{a}{2.1} \:=\:\frac{4.2-a}{3}\quad\Rightarrow\quad a\:=\:\frac{147}{85}\;\;{\color{blue}[1]}$

The distance is: .$\displaystyle AC + CE \;=\;\sqrt{(4.2-a)^2 + 3^2}+ \sqrt{a^2+ 2.1^2}\;\;{\color{blue}][2]}$

I'll let you substitute [1] into [2] . . .

• Mar 3rd 2008, 06:10 PM
rcmango
Yes, that is how I remember it, there is a distance that is known, then you subtract the unknown distance from it, and then solve for the unknown variable. I like setting problems up that way, appreciate the nice diagram to.

Also, I really appreciate all the extra help i got for this problem, i can see whats going on from different views. Awesome job. Thankyou.