Image Distance with two concave lenses

AI Thread Summary
The discussion revolves around calculating the final image location formed by a system of two lenses: a convex lens followed by a concave lens. The first step involves using the thin-lens equation to find the image distance from the convex lens, which is initially miscalculated due to sign errors. It is clarified that the image formed by the first lens serves as a virtual object for the second lens, requiring careful attention to the distances and signs when applying the thin-lens equation again. The final image location depends on correctly interpreting the virtual object distance for the concave lens. Proper understanding of lens behavior and sign conventions is crucial for solving the problem accurately.
Jende
Messages
4
Reaction score
0

Homework Statement


An object 1.25 cm tall is placed 100 cm in front of a convex lens with a focal length of magnitude 50 cm. A concave lens with a focal length of magnitude 20 cm is placed 90 cm beyond the first lens. Where is the final image located?

Homework Equations


(1/s)+(1/s')=1/f
s= object distance from lens
s'= image distance from lens
f= focal length

The Attempt at a Solution


I wasn't sure where to start so I tried find the image distance for the first lens.
(1/100cm)+(1/s')=(1/50cm) => (1/100cm)-(1/50cm)=(-1/100cm)=(1/s') => s'= -100

After this i got stuck because I wasn't sure what to do when the image distance is behind the second lens.

Thanks in advance.
 
Physics news on Phys.org
Jende said:

Homework Statement


An object 1.25 cm tall is placed 100 cm in front of a convex lens with a focal length of magnitude 50 cm. A concave lens with a focal length of magnitude 20 cm is placed 90 cm beyond the first lens. Where is the final image located?

Homework Equations


(1/s)+(1/s')=1/f
s= object distance from lens
s'= image distance from lens
f= focal length

The Attempt at a Solution


I wasn't sure where to start so I tried find the image distance for the first lens.
(1/100cm)+(1/s')=(1/50cm) => (1/100cm)-(1/50cm)=(-1/100cm)=(-1/s') => s'= +100
You have sign errors here.
Jende said:
After this i got stuck because I wasn't sure what to do when the image distance is behind the second lens.

Thanks in advance.
Take the object distance negative.
 
  • Like
Likes Jende
The image formed by the first lens acts as the object for the second lens. However, when you apply the thin-lens equation for the second lens you'll need to make sure that the object distance is the distance measured from the second lens, which you can do by taking into account the distance between the two lenses.

Be careful of the signs. As ehild says, you have an error in your first calculation.
 
  • Like
Likes Jende
Being behind he second lens this image is then a virtual object for the second lens. The light rays are still entering the second lens, but they seem to originate from a object located on the "wrong" side of the lens. The rays will therefore not form the object and is thus a virtual object for the second lens.
 
  • Like
Likes Jende
Ok thanks a lot. This was giving me a lot of frustration.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
View full post »

Thread 'Egg Drop Challenge! Need ideas for winning designs'

I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
View full post »

Similar threads

Focal length of the eyepiece lens in a compound microscope
Replies
7
Views
1K
Analyzing this 2-lens optical system
Replies
5
Views
1K
Find the location and length of the final image
Replies
2
Views
884
Inserting thick lenses into a thin lens system and deducing values
Replies
1
Views
2K
Geometric optics: Thin lense equation
Replies
3
Views
2K
Calculate a radius of a circle on a screen
Replies
3
Views
1K
Final Image position for a setup with two lenses
Replies
26
Views
4K
Mirror placed horizontally between two half convex lenses
Replies
11
Views
1K
Image position in an optical system?
Replies
5
Views
2K
Optics Problem with a Double Lens System
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top