Finding Convex Lens Focal Length

[dh743]

Homework Statement

Following data recorded:

Di (distance image to lens) 6.0 8.0 10.0 12.0 14.0
Magnification 0.2 0.6 1.0 1.4 1.8

Graph magnification against d_i and use the graph to find the focal length of the lens.

Homework Equations

\frac{1}{di}+\frac{1}{d(object)]}=\frac{1}{f}
M=\frac{di}{do}

The Attempt at a Solution

I worked out that the gradient should be equal to \frac{M}{di}
=\frac{1}{do}

Since the gradient is obviously 0.2, this leaves me with a constant value for do of 5cm, which is in fact the correct answer for the focal length. It doesn't make sense that there would be constant do value, nor does it make sense that this value would be the same as the focal length since no image is formed when object on focal length. Any instruction on manipulating these formulae is much appreciated.

 

[rl.bhat]

Th lens formula is
1/di + 1/do = 1/f

1 + di/do = di/f or
1 + M = di/f. or M = di/f - 1
It is in the form
y = mx + c. Here y is M, x is di and m is 1/f.

 

[dh743]

Th lens formula is
1/di + 1/do = 1/f

1 + di/do = di/f or
1 + M = di/f. or M = di/f - 1
It is in the form
y = mx + c. Here y is M, x is di and m is 1/f.

Thanks for the reply. So you've multiplied everything by di, but what does that actually show? As in why would you do it? Also, why isn't the gradient simply 1/do since (di/do)/di = 1/do?

 

[rl.bhat]

You have collected data by keeping the screen at 6.0 8.0 10.0 12.0 14.0 and adjusting the position of the object to get clear image. In that position you have measured the magnification.
so you have drawn the graph of image distance vs magnification.
The gradient is (M1 - M2)/di1 - di2)
The gradient is 1/f, not 1/do. In the problem do is not constant.
To draw the graph, you have to form the relation between M, di and f by using the lens formula.