Fix a metre rule on the bench with the graduated face up. Place the illuminated object at the zero end of the rule and the screen at the other end as illust...

Answer Details

(a) Lens displacement (conjugate-foci) method

For each fixed object-to-screen distance \(D\), the lens is moved to the two positions \(x_1\) and \(x_2\) that each give a sharp image on the screen. The separation of these positions is \(L = x_1 - x_2\). The readings and evaluated quantities are tabulated below.

\(D\) (cm)	\(D^{2}\) (cm\(^{2}\))	\(x_1\) (cm)	\(x_2\) (cm)	\(L=x_1-x_2\) (cm)	\(L^{2}\) (cm\(^{2}\))	\(D^{2}-L^{2}\) (cm\(^{2}\))
100	10000	80.50	18.30	62.20	3868.84	6131.16
90	8100	70.20	19.00	51.20	2621.44	5478.56
80	6400	59.10	20.00	39.10	1528.81	4871.19
70	4900	46.70	22.00	24.70	610.09	4289.91
60	3600	29.00	7.60	21.40	457.96	3142.04

Sample evaluation for the first reading (\(D = 100\ \text{cm}\)):

\[ D^{2} = 100^{2} = 10000\ \text{cm}^{2}, \quad L = 80.50 - 18.30 = 62.20\ \text{cm} \]\[ L^{2} = 62.20^{2} = 3868.84\ \text{cm}^{2}, \quad D^{2}-L^{2} = 10000 - 3868.84 = 6131.16\ \text{cm}^{2} \]

Graph of \(D^{2}-L^{2}\) against \(D\)

Plotting \(\left(D^{2}-L^{2}\right)\) on the vertical axis against \(D\) on the horizontal axis gives a straight line through the origin:

Slope and value of K

Reading the slope from the line of best fit using a large triangle, taking two clear points on the line \((D = 100\ \text{cm},\ D^{2}-L^{2} = 6000\ \text{cm}^{2})\) and \((D = 50\ \text{cm},\ D^{2}-L^{2} = 3000\ \text{cm}^{2})\):

\[ S = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6000 - 3000}{100 - 50} = \frac{3000}{50} = 60\ \text{cm} \]\[ K = \frac{S}{4} = \frac{60}{4} = 15\ \text{cm} \]

(This \(K\) is the focal length of the converging lens, since the displacement method gives \(f = \dfrac{D^{2}-L^{2}}{4D}\), so a graph of \(D^{2}-L^{2}\) against \(D\) has slope \(4f\) and \(K = \tfrac{S}{4} = f\).)

Two precautions

1. I avoided parallax error by reading the position of the lens on the metre rule with the line of sight directly above the mark.
2. I ensured that a sharp, well-focused image was formed on the screen before taking each reading of the lens position.

(b)(i) Real image versus virtual image

Real image	Virtual image
Formed by the actual intersection of refracted rays.	Formed where the refracted rays only appear to meet when produced backwards.
Can be caught (focused) on a screen.	Cannot be caught on a screen.
Inverted for a single converging lens.	Upright.

(b)(ii) Ray diagram: converging lens forming a real, diminished image

With the object placed beyond \(2F\), the converging lens forms a real, inverted and diminished image between \(F\) and \(2F\) on the far side of the lens.