Solve Convex Lens Question: Find Focal Length & Magnification

[toolcake]

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

 

[denverdoc]

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.

 

[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

$$\frac{80}{80O - O^2} = \frac{80}{(O + 16)(64 - O)}$$

 

[toolcake]

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!

Thanks for your help :D

 

[denverdoc]

In one case I=32, O=48 and in the other vice versa. I did the math as I posted, and then the light went on! So it didn't need to be covered in advance, but the results from doing the math show what might have been gathered from careful inspection of the problem...

 

[andrevdh]

Well, two facts:

I + O = 80

and f needs stay the same in both cases. Whether one puts O = 48 and I = 32 or O = 32 and I = 48, f will come out the same value plus the sum of I and O stays the same.

The magnification will be different in the two cases, the second magnification will be the inverse of the first. The first magnification will be > 1.0 and the second < 1.0

 

[toolcake]

Well, two facts:

I + O = 80

and f needs stay the same in both cases. Whether one puts O = 48 and I = 32 or O = 32 and I = 48, f will come out the same value plus the sum of I and O stays the same.

The magnification will be different in the two cases, the second magnification will be the inverse of the first. The first magnification will be > 1.0 and the second < 1.0

Thanks I get it now