TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with an illuminated object, converging lens, screen, metre rule, and other necessary materials. Measur...

Size of the illuminated object: \(a_{o} = 1.5\ \text{cm}\).

For each value of the object distance \(U\) the screen is adjusted until a sharp image is formed, the image size \(a\) is measured, and the linear magnification \(m=\dfrac{a}{a_{o}}\) together with its reciprocal \(m^{-1}\) are evaluated.

Table of readings

\(U\ /\ \text{cm}\)	\(a\ /\ \text{cm}\)	\(m=\dfrac{a}{a_{o}}\)	\(m^{-1}\)
30.0	2.2	1.467	0.682
35.0	1.5	1.000	1.000
40.0	1.2	0.800	1.250
45.0	0.9	0.600	1.667
50.0	0.7	0.467	2.143

Graph of \(m^{-1}\) against \(U\)

The points are plotted with \(m^{-1}\) on the vertical axis and \(U\) on the horizontal axis, and a straight line of best fit is drawn through them.

Slope of the graph

Using two widely separated points on the line of best fit, \((U_{1},\,m^{-1}_{1})=(30.0,\,0.63)\) and \((U_{2},\,m^{-1}_{2})=(50.0,\,2.07)\):

\[ s=\frac{\Delta(m^{-1})}{\Delta U}=\frac{2.07-0.63}{50.0-30.0}=\frac{1.44}{20.0}=0.072\ \text{cm}^{-1}. \]

Intercept on the vertical axis

Extending the line of best fit to \(U=0\):

\[ C=m^{-1}-sU=0.63-(0.072\times30.0)=0.63-2.16=-1.53. \]

So the intercept on the vertical axis is \(C\approx-1.5\).

Value of \(U\) for which \(m^{-1}=0\)

The line is \(m^{-1}=sU+C=0.072U-1.53\). Setting \(m^{-1}=0\):

\[ 0=0.072U-1.53\ \Rightarrow\ U=\frac{1.53}{0.072}=21.3\ \text{cm}. \]

Two precautions

• I avoided parallax error by reading the metre rule and the image size with the line of sight perpendicular to the scale.
• I ensured the image was as sharp (well focused) as possible before recording each image size.

(b) i. Value of \(m\) when \(U=37\ \text{cm}\)

From the line of best fit, at \(U=37\ \text{cm}\):

\[ m^{-1}=0.072(37)-1.53=2.66-1.53=1.13. \]\[ m=\frac{1}{m^{-1}}=\frac{1}{1.13}=0.88. \]

Therefore \(m\approx0.88\).

(b) ii. Real, diminished image formed by a converging lens

When the object is placed beyond \(2F\), the converging lens forms an image between \(F\) and \(2F\) on the far side that is real, inverted and diminished.