Analyzing this 2-lens optical system

[malolapjk]

Homework Statement

I have a two-lens system. Both principal axis coinciding. Separated by a distance 20cm. Both have focal length 15 cm. Will this act as a converging system or diverging system? If I consider a parallel beam of light is incident on it and I solve by first principles taking lens by lens I get a diverging system. A virtual image is formed between the two lens. But if I use the formula 1/f = (1/f1)+(1/f2)-(d/f1*f2), where fs are the focal lengths and d the separation between them I get a positive focal length. Which means it is a converging system forming a real image. How is that?

[b][LaTeX edited by a Mentor][/b]

Relevant Equations

1/f = (1/f1)+(1/f2)-(d/f1*f2)

• Using the Lens Formula:
  • The lens formula $1f=1f1+1f2−df1f2\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}f1=f11+f21−f1f2d$ is applicable for finding the effective focal length of the compound lens system.
  • Here, $f1=f2=15f_1 = f_2 = 15f1=f2=15 cm$ (focal lengths of each lens), and $d=20d = 20d=20 cm$ (distance between the lenses).
• Calculation:
  • Substitute the values into the formula: $1f=115+115−2015×15\frac{1}{f} = \frac{1}{15} + \frac{1}{15} - \frac{20}{15 \times 15}f1=151+151−15×1520$.
  • Simplify to find $1f=215−20225\frac{1}{f} = \frac{2}{15} - \frac{20}{225}f1=152−22520$.
  • $1f=30−20225\frac{1}{f} = \frac{30 - 20}{225}f1=22530−20$.
  • $1f=10225\frac{1}{f} = \frac{10}{225}f1=22510$.
  • $f=22510f = \frac{225}{10}f=10225$.
  • $f=22.5f = 22.5f=22.5 cm$.

• Lens-by-Lens Analysis​

  Lens 1:
  
  1. Parallel rays enter the first converging lens with $f1=15f_1 = 15f1=15 cm$.
  2. These rays converge to a point 15 cm beyond the first lens.
     Between Lens 1 and Lens 2:
  1. The distance between the two lenses is 20 cm.
  2. Thus, the rays would converge 5 cm before reaching the second lens.
     Lens 2:
  1. The rays enter the second converging lens 5 cm after they would have converged if uninterrupted.
  2. The object for Lens 2 is virtual and located 5 cm on the object side (left side) of Lens 2.
  3. For Lens 2, the object distance (uuu) is −5-5−5 cm (negative because it is virtual).
     Using the lens formula for Lens 2:
  
  $1f2=1v2−1u2\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}f21=v21−u21$
  where:
  
  • $f2=15f_2 = 15f2=15 cm$
  • $u2=−5u_2 = -5u2=−5 cm$
    $115=1v2+15\frac{1}{15} = \frac{1}{v_2} + \frac{1}{5}151=v21+51$
  Solving for v2v_2v2:
  
  $1v2=115−15=115−315=−215\frac{1}{v_2} = \frac{1}{15} - \frac{1}{5} = \frac{1}{15} - \frac{3}{15} = -\frac{2}{15}$
  
  $v21=151−51=151−153=−152v2=−152=−7.5 cmv_2 = -\frac{15}{2} = -7.5 \, \text{cm}v2=−215=−7.5cm$
  
  The negative value of v2v_2v2 indicates that the image formed by the second lens is virtual and located 7.5 cm on the object side (left side) of Lens 2. Therefore, the emergent rays are diverging after passing through Lens 2.

 

[kuruman]

Please fix the LaTeX code to make legible. Click on the link "LaTeX Guide", lower left above "Attach files" to learn how.

 

[WWGD]

It seems it's mostly about wrapping the text in question with ##'s.

 

[berkeman]

Please fix the LaTeX code to make legible. Click on the link "LaTeX Guide", lower left above "Attach files" to learn how.

It seems it's mostly about wrapping the text in question with ##'s.

I edited the LaTeX the best that I could, but there were still some terms that did not make sense to me. Hopefully the OP can chime in with any fixes.

@malolapjk -- I sent you a message with some tips about using LaTeX, in addition to the "LaTeX Guide" link below the edit window to the left.

 

[Steve4Physics]

Looks like the OP isn't going to reply. But for information...

Relevant Equations: 1/f = (1/f1)+(1/f2)-(d/(f1*f2)) [edit: pair of brackets added]

From memory, the above equation applies only when $d \le f_1$ - which is not the case in this problem. So using the equation leads to an incorrect answer. Maybe someone knowledgeable in optics can confirm.

(Edit. If the problem is only to choose between 'converging and 'diverging', then a simple ray diagram is all that is needed.)

 

[kuruman]

If the problem is only to choose between 'converging and 'diverging', then a simple ray diagram is all that is needed.

Indeed. However, I think that the formula that OP presented without derivation is incorrect. Here is my derivation. I would use the thin lens equation in the form $$\frac{1}{o}+\frac{1}{i}=\frac{1}{f}$$ and put the "object" at infinity ($o_1=\infty$) so that the image after light passes through the first lens
$\dfrac{1}{i_1}=\dfrac{1}{f_1}\implies i_1=f_1.$
This image is an "object" at distance $o_2=d-f_1$ from the second lens. Then we can find the position of the final image $i_2$, $$\frac{1}{d-f_1}+\frac{1}{i_2}=\frac{1}{f_2}\implies \frac{1}{i_2}=\frac{1}{f_2}- \frac{1}{d-f_1}.$$Now for the composite lens of focal length $f_{comp}$ we have an object at infinity forming an image at distance $i_2$ from the second lens. It follows that $$\frac{1}{f_{comp}}=\frac{1}{i_2}=\frac{1}{f_2}- \frac{1}{d-f_1}\implies f_{comp}=\frac{f_2(d-f_1)}{d-(f_1+f_2)}.$$ Notes

1. This derivation assumes that the focal lens of the composite system is defined as the distance from the last element of the system to the point where the exiting rays (or their extensions if diverging) cross the optical axis.
2. When $d< f_1+f_2$, as it the case here, $~f_{comp}<0$ which means that the composite lens is diverging. That's because the real image formed by lens 1 is less that one focal length of lens 2 which means that the final image is virtual.
3. When $d< f_1+f_2~$ and $~d< f_1$, $~f_{comp}>0$ which means that the composite lens is converging. That's because the image formed by lens 1 is formed on the "other" side of the two lenses. This makes it a "virtual object" for lens 2 which means that $o_2$ in the derivation above must change sign. The double change in sign means that the composite lens is converging.
4. When the two thin lenses are butting against each other $(d=0)$, the expression reduces to a converging lens of focal length$$f_{comp}=\frac{f_1f_2}{f_1+f_2}.$$

I agree that the OP may not resurface. This is to complete the picture.