# Concave mirror equation calculation

## Homework Statement

A physics student wants toobtain a real and virtual image that each is 3 times as large as the original object she has a concave mirror with a 0.50 m radius of curvature. what object distance will she require to obtain the real image?

## Homework Equations

1/do + 1/di = 1/f

## The Attempt at a Solution

m = 3 = -di / do 3do = -di

1/d + 1/3d =1/0.25

[1/d + 1/3d =1/0.25]3d

2d + 1 = 3d/0.25

[2d + 1 = 3d/0.25]0.25

0.5d + 0.25 = 3d

2.5d= 0.25

d= .10m m =do huh? I know I did something wrong

based upon my ray diagram do= .32m ish
and if that was the do then there would not be a real image formed becaues the object is inside of the focal point so only a virtual image would be formed right?

since m = 3 = -di / do 3do = -di

I tried to make di negative and that didnt turn out either I am not sure what I have done wrong

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rl.bhat
Homework Helper
1/d + 1/3d =1/0.25
Write this step as
1/d + 1/3d = 4 and solve for d

after looking at the problem today I see that I subtracted d from 3d instead of divided. what a silly mistake lol o well thx anyways.