Two Lense Problem: Solving A, B & C

  • Thread starter Thread starter Jacob87411
  • Start date Start date
AI Thread Summary
To solve the two-lens problem, first calculate the image position from the first lens using the formula 1/f = 1/do + 1/di, treating the second lens as absent initially. The image produced by the first lens then serves as the object for the second lens, requiring a second application of the same formula. For part A, the image is located 10 cm in front of the second lens, which is 50 cm from the first lens. Part B involves determining the final image height based on the magnification from the first lens, while part C mirrors part A's approach but with the object placed 5 cm in front of the two lenses in contact. Using scale drawings can help visualize the distances and relationships between the lenses.
Jacob87411
Messages
170
Reaction score
1
A converging lens of focal length 20 cm is separated by 50 cm from a converging lens of focal length 5 cm. (A) Find the final positiom of the image of an object placed 40 cm in front of the first lens. (B) If the height of the object is 2 cm, what is the height of the final image. (C) Determine the image position of an object placed at 5 cm in front of the two lens in contact.

I am having some issues with two lense problems.

A)In A if the image is 40 cm infront of the first lens it is 10 from the second. The image will come up somewhere inbetween the two? I am assuming I need the equation 1/f=1/d0 + 1/di, but do I use it twice or what.

B) This one isn't that bad, once I find A this shouldn't be an issue.

C) Part C is sort of just like A, once I can figure out how to solve A this one shouldn't be that bad.
 
Physics news on Phys.org
Jacob87411 said:
A)In A if the image is 40 cm infront of the first lens it is 10 from the second. The image will come up somewhere inbetween the two? I am assuming I need the equation 1/f=1/d0 + 1/di, but do I use it twice or what.

Yes, you have to use it twice. First, pretend the second lens isn't there, and find the position of the image that the first lens produces. Second, pretend that the first lens isn't there any more, put the second lens back where it was, and pretend that the object is located at the image position that you calculated in the first step. In other words, the image produced by the first lens becomes the object for the second lens. It's a good idea to make a scale drawing to keep track of where everything is, and see how the distances fit together.
 
Alright thanks, that makes sense
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top