O-LevelPhysicsLight5054/42 · May/June 2023 · Q314 Marks

A student investigates the image formed by a converging lens. The student arranges the apparatus as shown in Fig. 3.1. triangular object lamp to power supply- H white screen converging lens u [Figure 3.1 (not to scale)] The illuminated triangular object is shown full size in Fig. 3.2. H [Figure 3.2]

(a)Measure and record the height H of the triangular object. H = cm [1][1]
(b)(i)The student: • switches on the lamp and places the lens a distance u = 20.0cm from the triangular object • adjusts the position of the screen until a sharp, focussed image of the triangular object is formed on the screen. The image is shown full-size in Fig. 3.3. h [Figure 3.3] Measure the height h of the image on the screen shown in Fig. 3.3 on page 11. h = [1][1]
(b)(ii)Calculate the value of 1 h Give your answer to 2 significant figures. 1 = [1] h Add your values for h and 1 h to Table 3.1. [1][1]
(c)(i)The student repeats (b) for different values of u. He records all his readings in Table 3.1. Table 3.1 u/cm h 20.0 25.0 3.9 30.0 2.6 40.0 1.6 50.0 1.1 1 h Complete the headers by adding appropriate units. [1][1]
(c)(ii)Calculate the remaining values of 1/h and add them to Table 3.1. [1][1]
(d)The screen is a square sheet of white card of side 10cm. Look at the data in Table 3.1 and suggest why the student does not use values of u which are less than 20.0cm. [1][1]
(e)On the grid provided, plot a graph of 1/h on the y-axis against u on the x-axis. Start both axes from the origin (0, 0). Draw the straight line of best fit. 0 0 [Figure provided for graph plotting] [4][4]
(f)(i)Calculate the gradient m of your line. Show all working and indicate on the graph the values you use. m = [2][2]
(f)(ii)Calculate the focal length f of the lens. Use your value of H from (a) and the equation: f = 1/(mH) f = cm [1][1]
(g)When measuring the height of the image on the screen, the student's hand and the ruler obstruct the light from the object and prevent it from reaching the screen. Suggest one improvement to the apparatus used by the student to overcome this problem. [1][1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

[1] (a) 2.6 ± 0.1 (cm);
[1] (b)(i) 7.8 ± 0.1 (cm);
[1] (b)(ii) 0.13 – answer given to 2 significant figures only, correctly rounded ;
[1] (c)(i) both unit headings correct ;
[1] (c)(ii) 0.13, 0.26, 0.38, 0.63, 0.91 ;
[1] (d) any one from: image too big to fit on screen ; object inside focal length ; image becomes virtual ;
[4] (e) axes labelled, quantity and unit ; scales linear, not awkward, start from (0,0); points plotted accurately, to the nearest ½ square ; thin best-fit straight line drawn ;
[2] (f)(i) values indicated on graph, ≥ 1½ the distance between the extreme plotted points ; m = 0.026 ± 0.004 ;
[1] (f)(ii) calculation of f correct from candidate's values ;
[1] (g) any one from: graph paper or a scale on the screen ; clamp ruler near/by the side of the screen ; translucent screen and measure at the back;