for focal lenth of convex lens in practical exam:
i cant understand how to do/. explain. please!
U-V Method:

Trial no.           position of lens           position of image           Focal length[ uv/u+v}
1.
2.
                                          Mean=

Distant object method:

trial no.          nature of image        object dist.             image dist.        focal lenth[uv/u+v}
                       u<2f magnified
1.
2.
                       u>2f dimnished
1.
2.
?
 

Dear Student,

Please find below the solution to the asked query:

The relation between u, v and f for a convex lens is,

$\frac{1}{V}-\frac{1}{U}=\frac{1}{f}$
Using the proper sign convention, u is negative, u negative, v and f are positive. So we have,

$\frac{1}{V}=\frac{1}{U}+\frac{1}{f} \Rightarrow V=\frac{Uf}{U+f}$
The diagram of U ^-1 and V ^-1 will be as shown in the figure.

Comparing with $y = mx + c$, graph between $V^{-1} \& U^{-1}$  is with negative slope –1 and intercept  $f^{-1}$ .The corresponding graph is as shown in figure.

The focal length of the lens can be calculated by measuring the coordinates of either of the points A, B and C.



The u versus v graph is a hyperbola as shown in figure. By measuring the coordinates of point C whose coordinates are (2f, 2f) we can calculate the focal length of the lens.

In experiment, we have to take the readings of image distance V for different value of object distance U and we have to plot a graph.

The table of answers is as mentioned by you.
 

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards