Contact lenses, focal length and uncorrected far point distance

AI Thread Summary
A nearsighted individual uses contacts with a focal length of -7.6 cm and has a far point distance of 8.7 m with glasses. The formula 1/F = 1/do + 1/di is applied to determine the uncorrected far point distance. The calculations indicate confusion regarding the image distance, which remains constant due to the eye's anatomy. The discussion emphasizes the need to establish equations for both object and image distances to clarify the relationship between the lenses used. Understanding these principles is crucial for solving the problem effectively.
danok
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1. A nearsighted person wears contacts with a focal length of -7.6cm. If this person's far point distance with her glasses is 8.7m, what is her uncorrected far point distance?



2. 1/F=1/do+1/di



3. 1/-7.6=1/do+1/870
=-7.534

I am completely lost here...please help. Thanks!
 
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So it's not just me then...
 


The image distance is constant in both cases (the length of the eyeball)
You have an unknown focal length lens (the eye) and a combintion of the same unknown lens and a known lens with a known object distance.

So write the equations for object, image distance and focal length for both cases.
 
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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