Specification reference Checklist questions

a Draw a labelled diagram of a suitable arrangement to carry out this experiment.
(2 marks)
b Describe the necessary procedure to obtain an accurate and reliable value for the wavelength of the laser light.
Your answer should include details of all the measurements and necessary calculations.







(4 marks)

Chapter 5 Exam Answers

Question	Answer	Marks	Guidance
1 a	It showed that light was a wave (rather than a particle) / wave nature (of light). 	1
1 b i	Single wavelength (or frequency) 	1
1 b ii	(Waves / source(s) have) constant phase difference. 	1
1 b iii	Any sensible precaution, for example, do not look into laser / do not point the laser at others / do not let (regular) reflections enter the eye / safety signs / suitable safety goggles 	1
1 c	 0.02(0)   ecf from calculation of fringe spacing  6.0  10^–7 m  ( 600 nm) ecf from calculation of fringe spacing	3
1 d	Maxima closer together  (Quotes equation and states that) spacing is proportional to wavelength / D and s are constant therefore as λ decreases so ω decreases  OR links smaller wavelength to smaller path difference 	2
	Total__9'>Total	9
From AQA Physics A PHYA2 Mechanics, Materials and Waves mark scheme January 2010 (Question 5)
2 a	n1 > n2  (Incident) angle > critical angle (allow θc not c) OR critical angle must be exceeded 	2	Allow correct reference to optical density. Allow nA > nB Do not allow: Angle passes the critical angle.
2 b		2	For second mark, do not allow 1.6  10^8 Allow: 1.66  10^8 or 1.70  10^8 Allow: 1.6̇  10^8
2 c	sin 72  1.80 sin θ  θ  31.895  31.9, answer is correct if more than two significant figures seen 	2	Correct answer on its own gets both marks. Do not allow 31 for second mark. Allow: 31.8–32
2 d	1.80 sin θc  1.40 OR θc  51.058  51.1°  (accept 51) OR 0.778 	2	Correct answer on its own gets both marks. Do not accept 50 by itself.
2 e	22 + their 2c (22 + 31.9  53.9)  53.9 > (51.1) critical angle  OR 2c + 22 < their 2d (θc), ecf from c and d  Angle less than critical angle 	2	If 2c + 22 > 5d then total internal reflection expected. If 2c + 22 < 5d then refraction expected. Allow max 1 for: Total__8'>Total internal reflection because angle > critical angle only if their 2d > 2c + 22.
	Total	10
From AQA Physics A PHYA2 Mechanics, Materials and Waves mark scheme June 2013 (Question 5)
3 a	One of: (spectral) analysis of light from stars (analyse) composition of stars Chemical analysis Measuring red shift / rotation of stars  Insufficient answers: ‘observe spectra’, ‘spectroscopy’, ‘view absorption / emission spectrum’, ‘compare spectra’, ‘look at light from stars’.	1	Allow: measuring wavelength or frequency from a named source of light. Allow any other legitimate application that specifies the source of light. E.g., absorption/emission spectra in stars, ‘observe spectra of materials’.
3 b i	first order beam first order spectrum first order image 	1	Allow ‘n=1’ , ‘1’ , ‘one’ , ‘1st’
3 b ii	The light at A will appear white (and at B there will be a spectrum) or greater intensity at A 	1
3 c	( d = 1 / (lines per mm × 10^3) = 6.757 × 10^–7 (m) or 6.757 × 10^–4 (mm)  ( n λ = d sin θ ) = 6.757 × 10^–7 × sin 51.0 ecf only for: Incorrect power of ten in otherwise correct calculation of d use of d = 1480, 1.48, 14.8 (etc) from incorrect order in 3 b ii = 5.25 × 10^–7 (m) ecf only for: incorrect power of ten in otherwise correct d from incorrect order in 3 b ii	3
3 d	n = d (sin 90) / λ or n = 6.757 × 10^–7 / 5.25 × 10^–7  ecf both numbers from 3 c = 1.29 so no more beams observed or answer consistent with their working OR 2 = d (sin θ) / λ or sin θ = 2 × 5.25 × 10^–7 / 6.757 × 10^–7  ecf both numbers from 3 c sin θ = 1.55 (so not possible to calculate angle) so no more beams  OR sin^–1 (2 × their λ) / their d )  (not possible to calculate) so no more beams ecf	2	Accept 1.28, 1.3 Second line gets both marks Conclusion consistent with working
	Total	8
From AQA Physics A PHYA2 Mechanics, Materials and Waves mark scheme June 2014 (Question 6)
4 a i	Use of sin c = n2 / n1  1.47 	2
4 a ii	Use of n1 sin θ1 = n2 sin θ2  Correct substitution  22.7 (°) 	3
4 b	Critical angle increases (closer to 90 degrees) due to n2/n1 being closer to 1  Angle at air–core boundary (θ1) will decrease due to larger critical angle at core–cladding boundary 	2
	Total	7
From AQA Physics B PHYB1 Harmony and Structure in the Universe January 2013 (Question 8)
5 a	Grating and screen shown with both labelled  Laser or laser beam labelled 	2
5 b	Correct use of (n)λ  d sin θ  Measure appropriate angle (for example, to first-order beam is the minimum required)  Method to measure angle (for example, tan θ  , spectrometer, accept protractor)  At least one way of improving accuracy / reliability  For full marks: also explain how d is calculated, for example, d  1 / lines per mm ( 103) 	4
	Total	6
From AQA Physics A PHYA2 Mechanics, Materials and Waves mark scheme January 2012 (Question 5)

Optical Fibres and Earthquakes – Wider Reading (Exam question 2019 AS Paper1)


E arthquakes are some of our planet's most devastating natural disasters. With almost no warning, an earthquake can level entire cities in a matter of minutes. For the millions of people living on or near fault lines, a few extra minutes of warning can mean the difference between life and death, so it's crucial that we're able to see major earthquakes coming as soon as possible.
Mostly, that requires building sensitive earthquake detectors that can measure the smallest of quakes, which can signal that a larger one is on the way. But standard earthquake detectors can only get us so far, which is why a group of Stanford researchers is turning to a different solution: fibre optic networks.
Fiber optic cables transmit information at nearly the speed of light, and are used by telecommunications companies around the world. But they're also used by oil and gas companies to monitor small quakes caused by drilling equipment. These companies exploit a property of the cables called 'backscatter' to track movement of the cables and record seismic events.
At one end of the cable is a laser, that shines light into the cable. Some of that light hits impurities in the glass walls of the cable and bounces back; this is the 'backscatter' referred to earlier. The signal that comes back down the cable can change depending on whether the part of the cable that caused the backscatter was still or moving, and recording those backscatter signals can give scientists a map of seismic activity across a large area.
Typically, these fiber optic detectors are mounted to the sides of pipelines and other equipment, to monitor them for damage. But to monitor earthquakes, the Stanford researchers needed to use unattached cables, which most people believed was impossible.
"People didn't believe this would work," said researcher Eileen Martin. "They always assumed that an uncoupled optical fiber would generate too much signal noise to be useful."
But using a 3-mile fiber loop on the Stanford campus, the researchers demonstrated that using optical fibers in this way is entirely possible. Using their fiber network, they've been able to detect around 800 seismic events, including the recent earthquake in Mexico and two small, local earthquakes measuring at 1.6 and 1.8 magnitude.
This means that scientists can detect earthquakes using the existing fiber optic cables that telecom companies have already laid down across the country. While these cables will never be able to match the sensitivity of traditional seismometers, they are significantly cheaper and can give a broader detection network.
"Every meter of optical fiber in our network acts like a sensor and costs less than a dollar to install," said researcher Biondo Biondi. "You will never be able to create a network using conventional seismometers with that kind of coverage, density and price."