Thin Lens & Images: Real, Inverted & Smaller Objects

AI Thread Summary
A thin converging lens forms a real image when the object is placed outside the focal length. It also produces an inverted image under the same condition. For the image to be smaller than the object, the object must be positioned outside twice the focal length. The responses provided in the discussion confirm the accuracy of these conclusions. This understanding is essential for solving related optics problems effectively.
Glen Maverick
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Homework Statement



Under what conditions will a thin converging lens form:
a) a real image,
b) an inverted image,
and c) an image smaller than the object?

Homework Equations



none

The Attempt at a Solution



I have concluded that:
a) the object needs to be outside the focal length
b) the same
c) the objects needs to be outside twice the focal length.

I am not sure if the answers are right. Can you please check for me? Thank you!
 
Last edited:
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Your answers are correct.
 
Thank you to both Stanton and rl.bhat. I was looking for this problem. :)
 
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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}...
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