Physics - Optics

A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall which is a distance of 5.94 m to the right of the slide. The image is larger than the size of the slide by a factor of 82.0. How far is the slide from the lens? Is the image erect or inverted? What is the focal length of the lens? Is the lens converging or diverging?

I don't understand what I'm not doing correctly. I'm using the magnification equation to get a substitute for the variable s, and then sticking that into the thin lens equation and solving for the distance.

That is the correct method. Let me see your calculations. I suspect you messed up a sign.

(1/s)+(1/s')= (1/f)
(1/s)+(1/s')= .3367
(1/s')=.3367-(1/s)
s'=(1/(.3367-(1/s)))

putting that into the magnification equation ->

m=-(s'/s)
m= -(1/(.3367-(1/s)))/s
m= -(1/((.3367-(1/s))s)
m= -(1/(.3367s-1))
putting 82 in for m ->
(.3367s-1)=-(1/82)
.3367s=(81/82)
s=2.93 m



Where did you get f=1/.3367 ? That was not given.

M= - s'/s
s= -82s'
Aren't you given s as 5.94? So distance object is ..

Now calculate f

Ohh, I was taking R to be the 5.94m in the equation f=(R/2), which wasn't right. Guess I got a little bit ahead of myself and it looks like i overcomplicated that WAY too much. Thanks so much for your help.

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asked by Tifani

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