A student investigates the image of an object formed on a screen by a converging lens, as shown in Fig. 2.1. [Figure 2.1] The student measures the height h₀ of the object and the distance d from the lens to the screen. The height hᵢ of the image is measured as shown in Fig. 2.2. [Figure 2.2] The experiment is repeated for different values of d. It is suggested that hᵢ and d are related by the equation (d + t/2) / f = hᵢ/h₀ + 1 where f is a property of the lens called the focal length and t is the thickness of the lens.
- (a)A graph is plotted of hᵢ/h₀ on the y-axis against d on the x-axis. Determine expressions for the gradient and y-intercept.[2]
- (b)The value of h₀ is (2.4 ± 0.1) cm. Values of d and hᵢ are given in Table 2.1. [Table 2.1] Calculate and record values of hᵢ/h₀ in Table 2.1. Include the absolute uncertainties in hᵢ/h₀.[2]
- (c)(i)Plot a graph of hᵢ/h₀ against d/cm. Include error bars for hᵢ/h₀.[2]
- (c)(ii)Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled.[2]
- (c)(iii)Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.[2]
- (c)(iv)Determine the y-intercept of the line of best fit. Include the absolute uncertainty in your answer.[2]
- (d)(i)Using your answers to (a) and (c)(iii), determine the value of f. Include an appropriate unit and the absolute uncertainty in your answer.[3]
- (d)(ii)Using your answers to (a), (c)(iii) and (c)(iv), determine the value of t.[1]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- (a) gradient = 1/f
- (a) y-intercept = -t/(2f) - 1
- (b) hᵢ/h₀ values: 0.71 or 0.708
- (b) hᵢ/h₀ values: 0.79 or 0.792
- (b) hᵢ/h₀ values: 0.92 or 0.917
- (b) hᵢ/h₀ values: 1.1 or 1.08
- (b) hᵢ/h₀ values: 1.3 or 1.29
- (b) hᵢ/h₀ values: 1.5 or 1.50
- (b) Absolute uncertainties in hᵢ/h₀ from ± 0.07 to ± 0.1 (or ± 0.10 or ± 0.11).
- (c)(i) Six points plotted correctly. Must be accurate to nearest half a small square. Diameter of points must be less than half a small square.
- (c)(i) Error bars in hᵢ/h₀ plotted correctly. All error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.
- (c)(ii) Line of best fit drawn. Points must be balanced. Do not accept line from top to bottom point. Line must pass between (55, 0.75) and (56, 0.75) and between (77, 1.40) and (78, 1.40).
- (c)(ii) Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted.
- (c)(iii) Gradient determined with clear substitution of data points into Δy/Δx. Distance between data points must be at least half the length of the drawn line.
- (c)(iii) Gradient of worst acceptable line determined.
- (c)(iii) uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
- (c)(iv) y-intercept determined from substitution into y = mx + c.
- (c)(iv) y-intercept determined using gradient from worst acceptable line.
- (c)(iv) uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line or uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
- (c)(iv) No ECF from false origin method.
- (d)(i) f determined using gradient with correct substitution shown. f = 1 / gradient (c)(iii)
- (d)(i) f determined using gradient and given to two or three significant figures and correct SI unit shown with correct power of ten e.g. 33 cm or 0.33 m or 33.1 cm or 0.331 m
- (d)(i) Absolute uncertainty in f determined. absolute uncertainty in f = (Δgradient / gradient) * f
- (d)(ii) t determined using y-intercept and given to two or three significant figures. Correct substitution of numbers required. t = 2f × (y-intercept + 1) or t = (2 × (y-intercept + 1)) / gradient
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