You are provided with an illuminated object, converging lens, screen, metre rule, and other necessary materials. Measure and record the size a\(_{o}\) of th...

Answer Details

Practical: linear magnification of a converging lens

For each object distance \(U\) the object and screen are placed on opposite sides of the lens and the screen adjusted until a sharp image forms; the image size \(a\) is measured, then \(m = \dfrac{a}{a_{0}}\) and \(m^{-1}\) are evaluated. A specimen table (values illustrative):

Theory of the graph. For a real image, \(m = \dfrac{v}{u}\) and \(\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\). Combining, \(m = \dfrac{f}{u - f}\), so

\[ m^{-1} = \frac{u - f}{f} = \frac{1}{f}\,U - 1. \]

Hence a graph of \(m^{-1}\) (vertical) against \(U\) (horizontal) is a straight line of slope \(s = \dfrac{1}{f}\) and vertical intercept \(C = -1\). The value of \(U\) for which \(m^{-1} = 0\) is \(U = f\) (image formed at infinity).

Two precautions: ensure the object, lens and screen centres are at the same height and lie on a straight line; focus for the sharpest image and avoid parallax when reading the sizes.

(b)(i) Read from the straight-line graph the value of \(m^{-1}\) at \(U = 37\text{ cm}\), then \(m\) is its reciprocal (this reading comes from the candidate's own graph).

(b)(ii) To produce a real, diminished image, the object is placed beyond twice the focal length (\(u > 2f\)); the image then forms between \(F\) and \(2F\) on the other side, real, inverted and smaller than the object.