Oxford Revise AQA A Level Physics | Chapter 23 answers

Chapter 23: Physics of the eye and ear

Question	Answers	Extra information	Mark	AO Spec reference
01.1	Converging As diverging lenses only produce virtual images/would not be able to capture an image for the diverging lens		1 1	AO1 3.10.1.2
01.2	Max 2 marks from: Metre ruler fixed to the table Middle of lens combination marked clearly on the ruler Distance measured to centre of light bulb Screen clamped in place so that it is touching ruler		max 2	AO1 ATj PS1.1
01.3	Complete line of best fit and use of intercept 15 D		1 1	AO2 MS3.4 3.10.1.1
01.4	15 D = 20 D + power of diverging power of diverging = −5 D The gradient of the graph is 1/both intercepts are the same so confidence in result is high	Must be negative for mark	1 1	AO1 MS3.4 AO3 3.10.1.1
02.1	Eye lens too strong/eyeball too long Cannot be brought to focus on the retina/fovea		1 1	AO1 3.10.1.2
02.2	u = \(\infty\) or negative sign for image distance v = 26 − 2 = 24 cm \( \begin{aligned} & \frac{1}{f} = \frac{1}{\infty} + \left( -\frac{1}{0.24} \right)\\ & \frac{1}{f} =\,-\frac{1}{0.24} =\,-4.2\mathrm{\ D}\\ \end{aligned} \)		1 1 1	AO2 3.10.1.2
02.3	Image has to form 12 cm from eye/10 cm from lens \( \begin{aligned} & – \frac{1}{0.24} = \frac{1}{u} – \frac{1}{0.1}\\ & \frac{1}{u} = \frac{1}{0.1} – \frac{1}{0.24}\\ & u = 0.17\mathrm{\ m\ } = 17\mathrm{\ cm}\\ \end{aligned} \)	Possible e.c.f from 02.2 power of lens	1 1	AO2 3.10.1.2
02.4	Non-spherical cornea Image is focused in a given plane and out of focus in perpendicular plane Cylindrical lens		1 1 1	AO1 3.10.1.2
03.1	X is blind spot Y is fovea		1 1	AO1 3.10.1.1
03.2	None in X and Y Much higher numbers than cone Decreasing as move towards A and B		1 1 1	A01 3.10.1.1
03.3	Three curves labelled left to right: blue, green, red Green>red>blue Blue: 375 to 500 Green: 425 to 675 Red: 475 to 725	(All wavelength ranges ±30) All ranges correct for 1 mark	1 1 1	AO1 3.10.1.1
03.4	Bright light – cones used, dim light rods Cones have smaller diameter so resolution greater in bright light/resolution less in dim light Object has colour in bright light/no colour in dim light		1 1 1	AO1 3.10.1.1
04.1	To read the book, the lens must be thicker/more powerful than the board Ciliary muscles contracts This is called accommodation	Allow argument in reverse	1 1 1	AO1 3.10.1.2
04.2	Rays refracted at cornea Rays refracted (less) at lens Rays meeting beyond the fovea on the optical axis		1 1 1	AO1 3.10.1.2
04.3	Use of negative for image distance \( \begin{aligned} & \frac{1}{f} = \frac{1}{0.2} + \left( -\frac{1}{0.78} \right)\\ & \frac{1}{f} = P = + 3.7\mathrm{\ D}\\ \end{aligned} \)		1 1	AO2 3.10.1.2
04.4	Lens can be made thinner and achieve same refraction/lenses are lighter		1	AO3 3.10.1.2
05.1	Max 2 marks from: Human range of hearing is 20 Hz to 20 kHz on average for a young adult Speaker A does not have sufficient range of frequencies at max and min/losing higher frequencies Speaker B goes above range of human hearing		max 2	AO2 3.10.2.2
05.2	On average, people above 45 cannot hear above 12 kHz, so range of both speakers is suitable		1	AO2 3.10.2.3
05.3	Use of relative intensity level = 10 log₁₀\(\frac{I_1}{I_0}\) \( \begin{aligned} & \text{Max intensity level – average intensity level } = 10\log_{10} \frac{I_2}{I_0}- 10\mathrm{\ log}_{10} \frac{I_1}{I_0}\\ & 4\mathrm{\ dB\ } = 10\log \left( \frac{I_2}{I_0} – \frac{I_1}{I_0} \right)\\ & 0.4 = \log \frac{I_2}{I_1}\\ & \frac{I_2}{I_1} = 2.51\\ \end{aligned} \)	Need idea of difference in intensity level for 3 marks Simple 4 dB = 10 log\(\left( \frac{I_2}{I_1} \right)\) with no explanation gains 1 mark	1 1 1	AO2 3.10.2.2
05.4	dB is a logarithmic scale and so increases in powers of 10 Factor: \( \begin{aligned} & 10 = 10 \log\frac{I_2}{I_1}\\ & \frac{I_2}{I_1} = 10\\ \end{aligned} \) This is 4 times greater		1 1	AO3 3.10.2.2
06.1	A: malleus/hammer B: incus/anvil C: stapes/stirrup	All 3 correct 2 marks 1 error 1 mark	1 1	AO1 3.10.2.1
06.2	The small bones in the ear act as levers/increase by factor of 1.5 The area of the oval window is much smaller than the area of membrane; therefore, pressure increased/15 × smaller \(\text{Since }P = \frac{F}{A}\)	Allow force magnifiers	1 1 1	AO1 3.10.2.1
06.3	Max 3 marks from: Helical spiral-shaped cavity filled with fluid Contains three membranes Covered in rows of hairs Different frequencies detected at different parts of cochlea/high frequencies base/low frequencies apex		max 3	AO1 3.10.2.1
06.4	\( \begin{aligned} & 60\mathrm{\ dB\ } = 10 \log_{10} \frac{I_1}{I_0}\\ & 0.6 = \log_{10} \frac{I_1}{1 \times 10^{-12}}\\ & I_{1} = 4.0 (3.98) \times 10^{-12}\\ & \text{W m}^{-2}\\ \end{aligned} \)		1 1 1	AO2 3.10.2.2
07.1	The threshold of (normal) human hearing		1	3.10.2.2 A02
07.2	Halfway between 30 and 30 phon curve 25 phon ±2		1	3.10.2.2 A03
07.3	Max 4 marks from: 40 phon loss: Recognition that conversation above 40 phon so can hear Conversation will be quiet but should be able to follow They may need to speak up 60 phon loss: Normal conversation falls below 60 phon curve for all but the lowest frequencies would struggle to hear unless much louder Would struggle to hear women’s voices	Answer should reference not just loss but difficulty with conversations for both hearing losses	max 4	3.10.2.2 A03
07.4	A peak rise in intensity levels for 4000 Hz		1	3.10.2.2 AO1
08.1	Noise damage can have a cumulative effect/exposure every day can cause damage A single loud noise can instantly cause damage (owtte)		1 1	3.10.2.3 AO3
08.2	Difference in intensity levels \(= 10 \log_{10} \frac{I_2}{I_0}- 10 \log_{10} \frac{I_1}{I_0}\) \( \begin{aligned} & 140 – 135 = 10 \log_{10}\left( \frac{I_2}{I_0} – \frac{I_1}{I_0} \right)\\ & 0.5 = \log_{10}\frac{I_2}{I_1}\\ & \frac{I_2}{I_1} = 3.2\\ & 3.2 \times \mathrm{\ greater}\\ \end{aligned} \)		1 1	3.10.2.2 AO2
08.3	\( \begin{aligned} \text{Intensity level } & = 10 \log_{10} \frac{I_1}{I_0}\\ & = 10 \log_{10} \frac{0.1}{1 \times 10^{-12}}\\ & = 110\mathrm{\ dB}\\ \end{aligned} \)		1 1	3.10.2.2 AO2
08.4	\( \begin{aligned} & \text{Use of } P \propto \frac{1}{r^2}\\ & Pr^{2} = \text{ constant}\\ & 0.1 \times 1^{2} = P \times 5^{2}\\ & P = \frac{0.1}{25} = 4 \times 10^{-3}\mathrm{\ W}\\ \end{aligned} \) \( \begin{aligned} \text{Intensity level } & = 10 \log_{10} \frac{4 \times 10^{-3}}{1 \times 10^{-12}}\\ & = 96\mathrm{\ dB}\\ \end{aligned} \) Yes, they should wear ear defenders		1 1 1 1	3.10.2.2 AO2

Skills box answers

Question	Answer
1	\( \begin{aligned} & \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{0.26}\,-\,\frac{1}{0.25}.\\ & \frac{1}{f} = -\,0.15\mathrm{\ D}\\ \end{aligned} \)
2	\( \frac{1}{f} = 2.3\ D; \frac{1}{v} = \frac{1}{f}\,-\,\frac{1}{u} = 2.3\,-\,\frac{1}{0.25};\,\frac{1}{v}\,=\,-\,1.7, \text{ so }v = \frac{1}{-1.7} = -\,0.59\mathrm{\ m}\). This is a virtual image (v is negative).
3	In order for the image to be seen, it must be formed at the unaided near point and be virtual. \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{0.25}\,-\,\frac{1}{0.65} = 2.5\mathrm{\ D}\)

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