1I think the question gives us the variables as
f=100 cm
u= to be calculated??
object height = 1km = 100,000 cm
image height = 20 cm
so magnification = v/u = 20/100000 = 2 x 10-4
In aerial mapping a camera uses a lens with a 100cm focal lenght.Find the height at which the airplane must fly,so that the photogrpah of a 1 km long strip on the ground fits excatly on the 20 cm long flimstrip of the camera.
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7 Answers
11so we have v=u x 2 x 10 -4
using the lenses formula
1/u + 1/v = 1/f
1/u (1+1/2x10-4) = 1/f
=> u = f x (1+1/2x10-4)
=> u = 100 x ( 1+5000 ) = 500100 cm = 5.001 KM
1Please check & tell if I have done any mistake in calcultions [1]
1I dont think that will make any difference.
Its just about following a sign convention [1]
but yes, the correct way to solve is by using lens formula ...
I did it but got the same answer. actually the negative sign gets cancelled & we get the same equation while calculating [3]
62Sushma.. an Advice. This might sound somewhat simple. But I guess you should be very careful with the sign convention. Dont mix them up. This might lead to a lot of confusion later. I strongly suggest not to try using/ learning both the conventions. I like the absolute frame convention far more. But i know guys who like the other one :)
That is ur take
Sign modifications in Sushma's post :-
Using proper sign conventions
f=100 cm
u= to be calculated??
object height = 1km = 100,000 cm
image height = 20 cm
so magnification = v/u = - 20/100000 = - 2 x 10-4
so we have v= - u x 2 x 10 -4
using the lenses formula
1/v - 1/u = 1/f
1/u (- 1/(2x10-4) -1) = 1/f
=> u = f x (- 1 - 1/2x10-4)
=> u = 100 x ( -1-5000 ) = -500100 cm = -5.001 KM
here Negative simply implies that the object is kept on left side of the Lens
