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## Homework Statement

An object is located 14.0cm in front of a convex mirror the image is being 7.00cm behind the mirror. A second object twice as tall as the first one is placed in front of the mirror, but at a different location. The image of this second object had the same height as the other imafe. how far in front of the mirror is the second object located?

I've seen the solution, where the do2 = +42.00cm

The last line of the solution is as follows and is where I get lost. (please see part 3 - attempt at solution)

## Homework Equations

1/do + 1/di = 1/f

hi/ho = -di/do

## The Attempt at a Solution

I've seen thttp://www.scribd.com/doc/27149366/Ch-25-PHYSICS-CUTNELL-SOLUTIONS-MANUEL-7th-EDITION-go-to-other-document-for-8th-version-conversion" [Broken], (problem 39) where the do2 = +42.00cm

The last line of the solution is as follows and is where I get lost

Using this result in the mirror equation, as applied to the second object, we find that

1/do + 1/di = 1/f

1/d0 = 1/f - 1/di

1/do = (1/-14.00) - (1/-0.25do)

...

do = 42.00cm

in the ... section this is my attempt which is incorrect.

1/do = (1/-14.00) - (1/-0.25do)

1/do = -0.25do - -14.00/ 3.5do

do = 1/(13.75/3.5)

do = 0.255 [completely incorrect]

I know I am looking at a basic error here forgetting fractions or something very basic but cannot recall what it is . Can someone break it down?

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