Examstyle Questions: Electromagnetic Waves

Jon measures the focal length f of a convex lens. He repeats the measurement several times. The mean value of the measurement is 0.125 m. The range over which the measurements vary due to experimental uncertainty is ± 0.005 m.
Jon correctly records the final result with the equation
f = 0.125 ± 0.005 m.
a) The minimum value for f indicated by this equation is 0.120 m.
Write down the maximum value for f indicated by this equation.
maximum value for f = .................. m
(1 Mark)
b) Jon calculates the power P of the lens using the relationship
P = 1 f For the mean vvalue f = 0.125 m P = 8.00 D.
Calculate the maximum value of the power corresponding to the minimum value of the focal length 0.120 m. Consider sensible significant figures.
maximum power = .................... D
(1 Mark)
c) Complete the equation below to indicate the range of values within which the power can be expected to lie.
P = 8.0 ± ............................. D
(1 Mark)
(Marks available: 3)

Name Wavelength γ rays < 10^{10} m x rays < 10^{9 }m ultraviolet 7 x 10^{7} to 10^{9} m Visible Light 4 x 10^{7}  7 x 10^{7} m Infrared 7 x 10^{7}  10^{3} m Microwave 10^{3}  10^{1} m Radiowaves > 10^{1} m a) γ rays and x rays can have similar wavelength. How are they different?
(2 Marks)
b) Ultraviolet is much more likely to cause the release of photoelectrons from a metal than visible light. Explain why this is so.
(2 Marks)
c) Visble light is diffracted when it passes through a slit less than a millimetre wide. Microwaves can be diffracted by a slit a few centimetres wide. Why?
(2 Marks)
d) Electromangetic waves may be planepolarised. What does this prove about the waves?
(1 Mark)
(Marks available: 7)
Answer outline and marking scheme for question:

a) 0.13 (0) m
(1 Mark)
b) 8.33 / 8.3
(1 Mark)
c) ± 0.3 / 0.4 D ecf on (b)
(1 Mark)
(Marks available: 3)

a) γ rays are emitted by the nuclei of some radioactive atoms as they lose energy (1 Mark). Xrays are emitted by electrons as they slow down and lose kinetic energy.(1 Mark)
(2 Marks)
b) Ultra Violet has a shorter wavelength and higher frequency than visible light. (1 Mark)
Because the energy carried by a photon E = hf, the uv photon carries more energy and so can release electrons from a metal more easily. (1 Mark)
(2 Marks)
c) Microwaves havve much larger wavelength (1 Mark) than visible light so are diffracted by longer objects. (1 Mark)
(2 Marks)
d) To be polarised, a wavemust be a transverse wave. So Electromagnetic waves are transverse waves.
(1 Mark)
(Marks available: 7)