## Convex Lens Question

An object and a screen are fixed at a distance of 80cm apart and a convex lens forms a real image of the object on the screen. When the lens is moved along its axis a distance of 16cm, a real image of the object is again formed on the screen. Find the focal length of the lens and the magnification in each case.

m=v/u
1/u + 1/v - 1/f

I've been at this for the last 20-30 minutes and I've come no closer to the solution. I've tried letting the first distance from the lens (u1) = 2f whereby the image will be the same size.. no help.. and I've tried letting the distance between the 2 focuses of the lens=16cm.. no help.. and I've tried letting u1=xcm and u2=16+xcm but again.. no help.. I'd appreciate if someone could assist me.Thanks

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 welcome to PF. There may be a better approach, been many moons since I looked at a lens problem. d1 and d2 must equal 80. hence, 1/d1-1/(80-d1)=1/f also since we have anothr real image formed after moving 16 cm, direction unspecified 1/(d1+16)-1/(80-16-d1)=1/f slogging thru the math gives a simple soln.
 Recognitions: Homework Help This problem is based on the fact that one gets the same answer for the focal length of a lens if the object and image distances are swopped around (which would keep their sum the same). That is if we have in the first instance object distance, O and image distance I the second setup gives O + 16 and I - 16. The object distance in the first case would therefore be the same as the image distance in the second case: O = I - 16. Together with I + O = 80 one can now easily solve the problem. Or if you like to work the maths you can go this way: $$\frac{1}{O} + \frac{1}{I} = \frac{I + O}{IO}$$ starting out with $$\frac{1}{O} + \frac{1}{80 - O} = \frac{1}{O + 16} + \frac{1}{64 - O}$$ and using the formula above to develop it further $$\frac{80}{80O - O^2} = \frac{80}{(O + 16)(64 - O)}$$

## Convex Lens Question

 Quote by andrevdh This problem is based on the fact that one gets the same answer for the focal length of a lens if the object and image distances are swopped around (which would keep their sum the same). That is if we have in the first instance object distance, O and image distance I the second setup gives O + 16 and I - 16. The object distance in the first case would therefore be the same as the image distance in the second case: O = I - 16. Together with I + O = 80 one can now easily solve the problem. Or if you like to work the maths you can go this way: $$\frac{1}{O} + \frac{1}{I} = \frac{I + O}{IO}$$ starting out with $$\frac{1}{O} + \frac{1}{80 - O} = \frac{1}{O + 16} + \frac{1}{64 - O}$$ and using the formula above to develop it further $$\frac{80}{80O - O^2} = \frac{80}{(O + 16)(64 - O)}$$
I've done what you advised and got out the right focal length as 19.2cm but I don't understan how the problem is based on the image-object swap thing seeing as that isn't covered in our textbook and seeing as the lens is moved not the object or screen. Also I'm confused as to why I also get the right answer for magnification when i take O=32cm and I=48cm.. I've got the problem out but I don't get the logic!!!