Physics

1. 1.

   A travelling wave has frequency 250 Hz and wavelength 1.20 m. Find its speed and period, to 3 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Frequency counts cycles per second and wavelength is distance per cycle; their product is distance per second.
   2. The reciprocal frequency gives the time of one cycle without using wavelength.

   Marking points

   • v = f lambda.
   • v = 300 m s^-1 (3 significant figures).
   • T = 1/f = 0.00400 s.

   Examiner tip: The period is measured in seconds, not hertz.

2. 2.

   Two points on a progressive wave are separated by 0.150 m along propagation. Wavelength is 0.600 m. Find the phase difference in radians and explain whether they are in antiphase.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The separation is one quarter of a wavelength, corresponding to one quarter of a full phase cycle.
   2. Antiphase would require half a wavelength or an odd multiple of it.

   Marking points

   • Phase difference = 2pi delta x/lambda.
   • The difference is pi/2 rad.
   • They are not in antiphase; antiphase requires pi rad modulo 2pi.

   Examiner tip: A quarter wavelength gives pi/2, not pi.

3. 3.

   A string fixed at both ends has length 0.800 m, tension 64.0 N and mass per length 0.0100 kg m^-1. Find its third-harmonic frequency and the number of antinodes. Assume uniform tension.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Three half-wavelengths fit into the string, so lambda3 = 2L/3.
   2. Use linear density, not the total string mass, in the speed equation.

   Marking points

   • Wave speed sqrt(T/mu) = 80.0 m s^-1.
   • f3 = 3v/(2L) = 150 Hz (3 significant figures).
   • There are 3 antinodes and nodes at both ends.

   Examiner tip: The third harmonic is not the third overtone.

4. 4.

   Coherent 600 nm light passes through slits 0.300 mm apart onto a screen 2.00 m away. Find fringe spacing and state the approximation and coherence requirement used.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Adjacent bright fringes differ in path difference by one wavelength.
   2. At small angles sin theta and tan theta are approximately y/D, yielding the linear spacing formula.

   Marking points

   • Use w = lambda D/d with lambda = 6.00e-7 m and d = 3.00e-4 m.
   • Spacing is 0.00400 m = 4.00 mm (3 significant figures).
   • Use small angles and a constant phase difference between same-frequency sources.

   Examiner tip: Convert both nano- and millimetres to metres before substituting.

5. 5.

   A grating has 500 lines per mm. Monochromatic light of wavelength 650 nm falls normally on it. Find the second-order diffraction angle and greatest possible order, to 3 significant figures where appropriate.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Line density must be inverted to obtain separation; 500 lines/mm means 500000 lines/m.
   2. An order must be an integer with |sin theta| <= 1; rounding 3.0769 up would demand an impossible sine.

   Marking points

   • Grating spacing d = 1/(500000) = 2.00e-6 m.
   • sin theta2 = 2lambda/d = 0.650, so theta2 = 40.5 degrees.
   • n <= d/lambda = 3.0769..., so the greatest order is 3.

   Examiner tip: Use arcsin for the angle and floor for the order limit.

6. 6.

   Unpolarised light of intensity 80.0 W m^-2 passes through ideal polarisers with axes at 0, 30 and 90 degrees to a fixed reference. Find the final intensity and compare with removing the middle polariser.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Each polariser changes the transmitted polarisation direction to its own axis.
   2. The last polariser sees light at 30 degrees, so its relative angle is 60, not 90; inserting the middle element allows nonzero transmission.

   Marking points

   • The first polariser transmits half: 40.0 W m^-2.
   • Apply successive relative angles: I = 40.0 cos^2(30 degrees) cos^2(60 degrees) = 7.50 W m^-2.
   • Without the middle polariser the crossed pair transmits zero ideally.

   Examiner tip: Malus' law uses the angle between consecutive axes, not each absolute angle.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.