DP Physics (last assessment 2024)

Question 18M.3.SL.TZ2.2

Date	May 2018	Marks available	[Maximum mark: 9]	Reference code	18M.3.SL.TZ2.2
Level	SL	Paper	3	Time zone	TZ2
Command term	Determine, Estimate, Explain, Show that	Question number	2	Adapted from	N/A

[Maximum mark: 9]

18M.3.SL.TZ2.2

A student carries out an experiment to determine the variation of intensity of the light with distance from a point light source. The light source is at the centre of a transparent spherical cover of radius C. The student measures the distance x from the surface of the cover to a sensor that measures the intensity I of the light.

M18/4/PHYSI/SP3/ENG/TZ2/02

The light source emits radiation with a constant power P and all of this radiation is transmitted through the cover. The relationship between I and x is given by

$I = \frac{P}{{4\pi {{(C + x)}^2}}}$

(a)

This relationship can also be written as follows.

$\frac{1}{{\sqrt I }} = Kx + KC$

Show that $K = 2\sqrt {\frac{\pi }{P}}$ .

[1]

Markscheme

combines the two equations to obtain result

«for example $\frac{1}{I}$ = K²(C + x)² = $\frac{{4\pi }}{P}$ (C + x)²»

reverse engineered solution – substitute K = $2\sqrt {\frac{\pi }{P}}$ into $\frac{1}{I}$ = K²(C + x)² to get I = $\frac{P}{{4\pi {{(C + x)}^2}}}$

There are many ways to answer the question, look for a combination of two equations to obtain the third one

[1 mark]

The student obtains a set of data and uses this to plot a graph of the variation of $\frac{1}{{\sqrt I }}$ with x.

(b.i)

Estimate C.

[2]

Markscheme

extrapolating line to cross x-axis / use of x-intercept

Use C = $\frac{{y{\text{ - intercept}}}}{{{\text{gradient}}}}$

use of gradient and one point, correctly substituted in one of the formulae

accept answers between 3.0 and 4.5 «cm»

Award [1 max] for negative answers

[2 marks]

(b.ii)

Determine P, to the correct number of significant figures including its unit.

[4]

Markscheme

ALTERNATIVE 1

Evidence of finding gradient using two points on the line at least 10 cm apart

Gradient found in range: 115–135 or 1.15–1.35

Using P = $\frac{{4\pi }}{{{K^2}}}$ to get value between 6.9 × 10^–4 and 9.5 × 10^–4 «W» and POT correct

Correct unit, W and answer to 1, 2 or 3 significant figures

ALTERNATIVE 2

Finds $I\left( {\frac{1}{{{y^2}}}} \right)$ from use of one point (x and y) on the line with x > 6 cm and C from(b)(i)to use in I = $\frac{P}{{4\pi {{(C + x)}^2}}}$ or $\frac{1}{{\sqrt I }}$ = Kx + KC