Finding fmax in Thin Lens Refraction Problem

AI Thread Summary
The discussion revolves around solving a lens refraction problem involving a light source, a lens with focal length f, and a screen at a fixed distance L. The goal is to demonstrate that there are two positions for the lens where an image can be formed on the screen, provided f is less than a maximum value, fmax. Participants suggest using the lens formula to derive a quadratic equation for the position of the lens, leading to two solutions for s. The conversation confirms the correct application of the quadratic formula and highlights the relationship between the lens position and image size. The problem remains unsolved in terms of finding fmax, as the lecturer has not provided answers.
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Homework Statement



Hi all. This is a problem from one of the problem sheets from earlier in the term. I'm just revising for my exam tomorrow and can't solve this one. Here it is:

Consider a light source placed at a fixed distance L from a screen, such that a lens of focal length
f can be placed between the source and the screen. Show that as long as f < fmax there are two
positions where the lens can be placed so that an image is formed on the screen, and find a value for fmax.

Homework Equations



1/s + 1/s' = 1/f

The Attempt at a Solution



I substituted in s' = L-s which gives me:

1/s + 1/(L-s) = 1/f

Given that there are two solutions and the nature of that formula I am guessing I need a quadratic equation and then to solve for s. Not sure how to expand that equation so I can solve for s?

Thanks
 
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multiply both sides by s(L-s) and you will get a quadratic equation.
 
So far so good. f and L are constants, right? Merely solve for s. Yep, you might a quadratic equation with two solutions.
 
Thanks for the help. Got the equation in the form (1/f)s^2 - (L/f)s + L = 0 giving the solutions s= (L±sqrt((L/f)^2 - (4L/f))/2

Does that look good? Lecturer hasn't published answers to this problem sheet for some reason. :(
 
your quadratic equation loo0ks good. I assume you are using the quadratic formula to solve for s. Just make sure you use the quad formula correctly. Note that when the lens is close to the object, the image is large, and when the lens is close to the screen, the image is small.
 
Yep I used the quadratic formula. Thanks for your help. :)
 
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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