How do I determine the distance between two lenses in a telephoto lens system?

AI Thread Summary
To determine the distance between two lenses in a telephoto lens system, the focal lengths of the objective and concave lenses are critical. The first lens creates an image that serves as the object for the second lens, necessitating careful calculation of distances using the lens formula. The user initially miscalculated the object distance for the second lens but corrected it by adjusting the sign in the formula. Understanding the virtual image concept is essential, as rays behave as if they originate from this virtual image. Mastering these calculations is crucial for successfully answering related exam questions.
creative_wind
Messages
7
Reaction score
0

Homework Statement



A simple camera telephoto lens consists of two lenses. The objective lens has a focal length f1 = +36.0 cm. Precisely 33.0 cm behind this lens is a concave lens with a focal length f2 = -10.0 cm. The object to be photographed is 3.67 m in front of the objective lens.

(a) How far behind the concave lens should the film be placed?

(b) What is the lateral magnification (ratio of image height to object height) of this lens combination?

Homework Equations



1/f=1/di + 1/do

M=-di/do

The Attempt at a Solution



My understanding is that the di1 (the objective lens) is the do2 (for the second/concave lens). Calculating di1 gives me 0.399 m which is greater than the distance between the two lenses, so that would mean do2 on the same side that the film would be on. I tried using 0.399m - 0.33 m = 0.069m = do2 to calculate di2 anyways, but that answer is wrong. I'm stuck as to what value to use for do2. As for part b) I know to find the magnification of each lens and then multiply the two magnification values to get the final magnification. I just need to figure out part a). Thanks so much for any help!
 
Last edited:
Physics news on Phys.org
creative_wind said:
My understanding is that the di1 (the objective lens) is the do2 (for the second/concave lens). Calculating di1 gives me 0.399 m which is greater than the distance between the two lenses, so that would mean do2 on the same side that the film would be on. I tried using 0.399m - 0.33 m = 0.069m = do2 to calculate di2 anyways, but that answer is wrong. I'm stuck as to what value to use for do2. As for part b) I know to find the magnification of each lens and then multiply the two magnification values to get the final magnification. I just need to figure out part a). Thanks so much for any help!

Hi creative_wind! :smile:

You really need to show us what you did if we're going to see where the mistake is.

I'll guess … did you remember to make the 1/f for the second lens negative? :smile:
 
Okay so as I wrote above I was using 0.399 m -0.33 m = 0.069 m. It should be 0.33 m - 0.399 m = -0.069 m, which gives me the correct answer. I can't say I really understand the concept of locating the position of the image of the first lens since it never really forms (hits second lens first) but at least I understand how to answer the question and could do it on an exam now. Thanks for your help!
 
… don't talk to me about reality …

creative_wind said:
I can't say I really understand the concept of locating the position of the image of the first lens since it never really forms …

You're really into reality aren't you? :smile:

Forget reality!

phsyics is reality … maths (including geometry) isn't :smile:

and this is a geometrical trick

the rays behave as if they did come from the virtual image …

just like the image in an ordinary flat mirror, which saves you the effort of having to draw incidence and reflection angles carefully!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top