Finding the distance in an optics system

[jisbon]

Homework Statement

An object is located 20cm to the left of a concave mirror (with curvature 10cm). Light leaves object is the first incident on the mirror and passes through the lens of 10cm in focal length located left of the object. A final image formed by the system is located a the same location as the original object.

i) Find distance between lens and mirror
ii) Comment on orientation and magnification of the final image

Relevant Equations

$\dfrac {1}{p}+\dfrac {1}{q}=\dfrac {1}{f}$

Hi all,
Pretty new to optics here, so would like to get my basics right.
Below is my working, was wondering if my concepts are correct here.

First I will separate mirror and lens and solve them 1 by 1, so solving for mirror first,

$\dfrac {1}{p_{1}}+\dfrac {1}{q_{1}}=\dfrac {1}{f_{1}}$ where $p_{1}$ is the object distance to mirror and $q_{1}$ is the image distance to mirror.
Putting $p_{1}$ as 20cm and ${f_{1}}$ as 5cm (since f=R/2) , I get $q_{1}$ to be 20/3 cm
From what I understand, since $q_{1}$ is positive, this means it is infront of the mirror, so it should be something like:

this?
Assuming this is correct,

Now moving on to the lens,
$\dfrac {1}{p_{2}}+\dfrac {1}{q_{2}}=\dfrac {1}{f_{2}}$
let x be the distance between between object and lens,
so from the above equation,
Taking $p_{2}$ = distance between object and lens
and
$q_{2}$ = distance between image and lens
$\dfrac {1}{p_{2}}+\dfrac {1}{q_{2}}=\dfrac {1}{f_{2}}$
where the new object is the image projected by the mirror,
$\dfrac {1}{x+13.333}+\dfrac {1}{x}=\dfrac {1}{-10}$
and from here I can find x where (i) will be x +20cm

Is this correct? Thanks

 

[Cutter Ketch]

That all looks right to me. On to part (ii)

 

[kuruman]

$\dfrac {1}{x+13.333}+\dfrac {1}{x}=\dfrac {1}{-10}$
and from here I can find x where (i) will be x +20cm
Is this correct? Thanks

The equation you have is correct, however it is quadratic in $x$ which means two solutions. Which one will you use and how?

On Edit: The equation is almost correct. You have to be careful about what $x$ is and how you use it in the equation. I would strongly recommend that you define $x$ as what you are looking for, namely the distance between the mirror and the lens. Then one of the solutions should come out positive and greater than 20 cm. The second solution should be something obvious in retrospect.

 

[jisbon]

The equation you have is correct, however it is quadratic in $x$ which means two solutions. Which one will you use and how?

On Edit: The equation is almost correct. You have to be careful about what $x$ is and how you use it in the equation. I would strongly recommend that you define $x$ as what you are looking for, namely the distance between the mirror and the lens. Then one of the solutions should come out positive and greater than 20 cm. The second solution should be something obvious in retrospect.

There are 2 solutions to x,

In this case taking x to be -28..68517 will make more sense, but I don't actually understand why in this case.
Following your edit:

$\dfrac {1}{x-\dfrac {20}{3}}+\dfrac {1}{x-20}=\dfrac {1}{-10}$ where x is the distance between mirror and lens
1st term: distance between obj and lens
2nd term: distance between image and lens
3rd term: focal length
I got:

which seems to be wrong since x must be more than 20cm?

 

[ehild]

There are 2 solutions to x, View attachment 255394
In this case taking x to be -28..68517 will make more sense, but I don't actually understand why in this case.
Following your edit:

$\dfrac {1}{x-\dfrac {20}{3}}+\dfrac {1}{x-20}=\dfrac {1}{-10}$ where x is the distance between mirror and lens
1st term: distance between obj and lens
2nd term: distance between image and lens
3rd term: focal length
I got: View attachment 255396

which seems to be wrong since x must be more than 20cm?

x-20 is the Physical distance between image and lens, but you need to use it with negative sign if the image is virtual.

 

[Cutter Ketch]

Oops! I didn’t catch it when I looked at it the first time. Sorry about that. In your final equation you write

$\frac {1} {x + 13.333} + \frac 1 x = \frac 1 {-10}$

but both the object and the image are on the same side of the lens. That means the image is virtual and the position should be negative

 

[jisbon]

Oops! I didn’t catch it when I looked at it the first time. Sorry about that. In your final equation you write

$\frac {1} {x + 13.333} + \frac 1 x = \frac 1 {-10}$

but both the object and the image are on the same side of the lens. That means the image is virtual and the position should be negative

Will the image always be virtual if the object and the image are on the same side of the lens (regardless of lens?/mirror even?)

 

[kuruman]

Will the image always be virtual if the object and the image are on the same side of the lens (regardless of lens?/mirror even?)

An image is virtual if it is formed where light has no business being relative to the object: behind a mirror and in front of a lens.

 

[Cutter Ketch]

Will the image always be virtual if the object and the image are on the same side of the lens (regardless of lens?/mirror even?)

Lens, yes. Mirror, vice versa.

 

[jisbon]

Ah yes okay, thanks for all your help :)

An image is virtual if it is formed where light has no business being relative to the object: behind a mirror and in front of a lens.

Lens, yes. Mirror, vice versa.