Mirror equations - basic math error - what is it?

[Hemingway]

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" , (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?

 

[ehild]

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]

Use parentheses correctly. 1/do = 1/(-14) - 1/(-0.25 do).

You can rewrite it as

1/do-1/(0.25do) =-1/14.

simplify the left-hand side, and take the reciprocal of both sides.

ehild

 

[collinsmark]

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

I'm not following you on that last step (in red).

Go back to the previous representation,

1/do = (1/-14.00) - (1/-0.25do)​

You have a d_o on the left side of the equation, and another d_o in one of the terms on the right side. Move that term on the right side to the left side of the equation. Then combine terms until you have a single term. Don't worry about the (1/-14.00) term until after you've combined all the terms with d_o.

[Edit: ehild beat me to the hint.]

 

[Hemingway]

1/do - 1/.25do = 1/-14
1/.75do = 1/-14
1.3do = 1/-14
...
This still doesn't seem to get to the answer.

Could someone show me step by step to how they would get to the answer? I am just too confused. I tried :) I'm finding it difficult to simplify with the inverse.

Cheers

 

[collinsmark]

1/do - 1/.25do = 1/-14
1/.75do = 1/-14

As ehild implied earlier, that's not the way to add fractions.

\frac{1}{a} + \frac{1}{b} \ne \frac{1}{a + b}

Instead, you need to find a common denominator.

So let's start with the equation you're working with:

\frac{1}{d_o} - \frac{1}{0.25 d_o} = \frac{1}{-14}

First find a common denominator, for the left side the equation. By inspection, notice that 0.25 goes into 1.0 four times. So an easy choice would be to multiply the numerator and denominator by 4 in the second term.

\frac{1}{d_o} - \frac{4}{(4)(0.25) d_o} = \frac{1}{-14}

\frac{1}{d_o} - \frac{4}{d_o} = \frac{1}{-14}

Now you can combine both terms on the left side of the equation:

\frac{1 \ - \ 4}{d_o} = \frac{1}{-14}

You should be able to finish it from there.
==============================================

There are other ways to find common denominators, so use whichever method you like best. But there is always one full-proof method that works in all situations, which is to multiply the denominators together. That will always be a common denominator (this method is useful if you can't think of a better common denominator first).

\frac{1}{a} + \frac{1}{b}

= \left( \frac{b}{b} \right) \left(\frac{1}{a} \right) + \left( \frac{a}{a} \right)\left(\frac{1}{b} \right)

= \frac{b}{ab} + \frac{a}{ab}

= \frac{a + b}{ab}

Thus,

\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}

Try this method on your particular problem and see if you get the same answer as the method used above.

 

[Hemingway]

Thank you! It's hard sometimes when you instinctively know you're doing the wrong thing - but you've just hit an intellectual wall. Thank you so much everybody for helping me out. Many thanks CollinsMark - that would have taken some time to do and I appreciate it.