Combining Lenses and Mirrors: A Short Guide to Calculating Focal Lengths

[Tanishq Nandan]

Homework Statement

Homework Equations

▪1/f=[(n2/n1)-1][1/R1 -1/R2]
▪1/f(eq)=1/f1 +1/f2
For combination of 2 lens
▪1/f(eq)=1/f(mirror) -2/f(lens)
for combination of lens and mirror
▪1/v+1/u=1/f
Mirror formula

The Attempt at a Solution

I just applied the above equations
1.Finding the individual focal lengths of the two concavoconvex lens using Eq1 (Ccoming as 200 and 400/3)
2.Finding their equivalent focal length using Eq2(coming as 80)
3.Finding equivalent of the lens and mirror using Eq3(coming -40/3).
4.Finally using the simple mirror formula,1/v+1/u=1/f,to find image distance(coming something which isn't the answer)
Is there anything wrong with this method or am I doing any calculation mistake.I MAY have done something wrong with the sign convention while applying Eq1,maybe..maybe not..
Wherever the mistake is,I would be glad if someone points it out.

 

[alejandromeira]

Puf, tremendous lens system !
You can try the following:
A) find the equivalent, lens1 + lens2 + mirror, as you have done.
B) with those diopters you must calculate the position of the image.
But now the rays of light go to the left, and they go back through the lenses to the left.
C) Use the position of the image calculated in B, as if it were an object, to calculate the position of the final image. Here you should only consider the power of the two lenses. (remember that when the light rays go to the left the focus of the lens system is changed of side).

AND I DON'T KNOW IF IT WILL WORK !

 

[Tanishq Nandan]

But,as soon as i finish A...I can replace it with a mirror of the calculated focal length.
Now,it's simple mirror formula.
So,what's the need for C?

 

[alejandromeira]

Because the light reflected in the mirror goes back through the lenses, from right to left.
You can try this. If your previous calculations are correct, it may work.
Personally, I would propose the problem differently.
But first try this, because this way, you already have some work done.

 

[Tanishq Nandan]

Personally, I would propose the problem differently

Please do.

 

[alejandromeira]

Ok. When I have a system of n lens, I apply the following principle: I place the center of figure on the first lens, and calculate its image.
I then place the center of figure on the second lens, and use the image of the first as if it were the position of the object. And I calculate the image formed by the second lens ... etc.
In this problem, the center of figure is always the same.

For your specific case I indicate the following procedure:

A) Calculate the diopters of lens assembly 1 and 2, so they work as a single lens.
B) Calculate the image formed by this lens assembly.
C) With the data of this image formed by the lenses, use them as if it were an object, and calculate with the mirror equation the position of the image formed by the mirror.

But the rays of the mirror now cross the lenses from right to left.

D) Then re-use the lens equation to calculate the final image. But now, as the beams of light go to the left, the focus are changed.

You have this other thread that is similar: https://www.physicsforums.com/threads/lens-reflector-final-idol-optics.923113/#post-5824779

Ok. I have helped you with everything I understand, but the problem is somewhat laborious.

 

[conscience]

Hello Tanishq ,

What is the official answer provided ? I have worked the problem .In case my answer matches , I will be in a position to help you .

 

[Tanishq Nandan]

It's 24cm to the left of the lens.

 

[conscience]

It's 24cm to the left of the lens.

OK. This is what I am getting .

Homework Statement

View attachment 210658

Homework Equations

▪1/f=[(n2/n1)-1][1/R1 -1/R2]
▪1/f(eq)=1/f1 +1/f2
For combination of 2 lens
▪1/f(eq)=1/f(mirror) -2/f(lens)
for combination of lens and mirror
▪1/v+1/u=1/f
Mirror formula

The Attempt at a Solution

I just applied the above equations
1.Finding the individual focal lengths of the two concavoconvex lens using Eq1 (Ccoming as 200 and 400/3)
2.Finding their equivalent focal length using Eq2(coming as 80)

OK till here .

▪1/f(eq)=1/f(mirror) -2/f(lens)
for combination of lens and mirror

3.Finding equivalent of the lens and mirror using Eq3(coming -40/3).

I think the problem lies here . I don't think that formula is valid .

You have correctly calculated the focal length of the two lenses . You need to treat the combined lens and mirror separately .

For the first refraction at the combined lens , what is the image distance ?

 

[Tanishq Nandan]

Umm...it's coming 240/23 to the left of the lens..but I doubt its correct..I might be making some mistake in sign convention...formula is 1/v-1/u=1/f...u will be negative,but will f??
If f is negative..it'll come 240/17 to the left.

 

[conscience]

Umm...it's coming 240/23 to the left of the lens..but I doubt its correct..I might be making some mistake in sign convention...formula is 1/v-1/u=1/f...u will be negative,but will f??
If f is negative..it'll come 240/17 to the left.

What value will you put in "u" in the lens formula ?

 

[Tanishq Nandan]

-12

 

[conscience]

OK . But that doesn't give v= 240/23 . Recheck your calculations .

 

[Tanishq Nandan]

What about f?? Should it be +80 or -80?
And 240/23...not 13

 

[conscience]

What about f?? Should it be +80 or -80?

+80

 

[Tanishq Nandan]

Then it's 240/17

 

[conscience]

Then it's 240/17

Yes . Now this image acts as an object for the mirror . Where is the image located for the reflection from the mirror ?

 

[Tanishq Nandan]

It's not coming 24,at least..
1/v+1/u=1/f
u is -240/17
f is -20
Anything wrong?

 

[conscience]

It's not coming 24,at least..
1/v+1/u=1/f
u is -240/17
f is -20
Anything wrong?

 

[Tanishq Nandan]

But concave mirrors have negative focal lengths..

 

[conscience]

But concave mirrors have negative focal lengths..

Yup .It will be negative but not 20 .

 

[Tanishq Nandan]

? how..why..
You're right..it should come 8..i think..but how??beats me..

 

[conscience]

? how..why..
You're right..it should come 8..i think..but how??beats me..

What is the relationship between focal length and radius of a spherical mirror ?

 

[Tanishq Nandan]

oooooooo..idiot..i was given radius.not focal length

 

[conscience]

oooooooo..idiot..i was given radius.not focal length

So ,what is the answer to question I asked in post#17 ?

 

[Tanishq Nandan]

Umm..it's coming 240/7..
Do i need to consider refraction from the lens system again?

 

[conscience]

Umm..it's coming 240/7..

Yup .

Now reflected rays from the mirror again suffer refraction from the combined lens .The image of the mirror acts as an object for the combined lens .

Note that , now the rays are traveling from right to left .

 

[Tanishq Nandan]

Yup..just did it..got the ans..thanks a tonne for bearing with me!

 

[alejandromeira]

Thaks, conscience. We have all learned a lot from you.

(my UTC does not allow me to participate more actively in the threads)