What is the Speed of the Image in Optics?

AI Thread Summary
An object is positioned 20 cm from a converging lens with a focal length of 5 cm, moving towards the lens at 12 cm/s. To determine the speed of the image moving away from the lens, the lens formula 1/f = 1/di + 1/do is applied, where both object distance (do) and image distance (di) are functions of time. The object's distance changes over time as do = 20 - 12t, leading to a negative velocity of -12 cm/s. Calculus is suggested to find the derivative of di at time t=0 when do equals 20 cm. This approach will yield the speed of the image as it moves away from the lens.
Lara_C
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Homework Statement



An object is placed 20cm from a converging lens of focal length 5cm. If the object moves towards the lens at a speed of 12cm/s, at what speed does the image move away from the lens?

Homework Equations



1/f=1/di+1/do
v=d/t

The Attempt at a Solution


Let's say the object moves...
Do=do1+do2
Di=di2-di1
so if di+fdo/do-f
vt=(fdo2/do2-f)-(fdo1/do1-f)
However, both vi and ti are unknown, so how should I proceed?
 
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Lara_C said:

Homework Statement



An object is placed 20cm from a converging lens of focal length 5cm. If the object moves towards the lens at a speed of 12cm/s, at what speed does the image move away from the lens?

Homework Equations



1/f=1/di+1/do
v=d/t

The Attempt at a Solution


Let's say the object moves...
Do=do1+do2
Di=di2-di1
so if di+fdo/do-f
vt=(fdo2/do2-f)-(fdo1/do1-f)
However, both vi and ti are unknown, so how should I proceed?

Hi Lara, welcome to PF.

Have you studied some Calculus? The object moves towards the lens so the object distance is do=20-12t in terms of time. The velocity of the object is d(do)/dt=-12 cm/s. The velocity of the image is vi=d(di)/dt. In the lens equation, both do and di are functions of the time. Find the derivative of di at time t=0 when do = 20 cm.

ehild
 
Thank you for your help!
 
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