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Transcript
IB PHYSICS UNIT 11 RESOLUTION
1.
(a) Using the axes below draw a graph to show how the intensity of the pattern varies with distance along the screen for
single slit diffraction. The point X on the screen is shown as a reference point. (This is a sketch graph; you do not need to
add any numerical values.)
(3)
X
(b)
In this experiment, the light from point source P has a wavelength of 500 nm and the width of the central maximum
of intensity on the screen is 10.0 mm. When light of unknown wavelength λ is used, the width of the central
maximum of intensity is 13.0 mm. Determine the value of λ.
(2)
Another point source Q, emitting light of the same wavelength as P (500 nm) is placed 5.0 mm from P and the two sources
are arranged as shown below.
P
Q
The distance between the sources and the slit is 1.50 m.
(c)
(i)
State the condition for the image of P and the image of Q formed on the screen to be just resolved.
(ii)
2.
(b)
Determine the minimum width b of the slit for the two images to be just resolved.
(a) State the name of the wave phenomenon that limits the resolution of any optical instrument.
Draw a diagram to show how the Rayleigh criteria are determined.
3.(a) Light from a point source is brought to a focus by a convex lens.
(i)
State why the image of the point source will not be a point image.
IB PHYSICS UNIT 11 RESOLUTION
(ii)
Describe the appearance of the image.
Two light receptors at the back of the eye are 4.0 μm apart. The distance of the receptors from the convex lens at the front
of the eye is 17.0 mm, as shown below.
4m
17 mm
Light of wavelength 550 nm from two point objects enters the eye. The centres of the images of the two objects are focused
on the light receptors.
(b)
(i)
Calculate the angle α in radians subtended by the two receptors at the centre of the eye lens.
(2)
(ii)
4.
Use the Rayleigh criterion to calculate the diameter of the pupil of the eye so that the two images are just
resolved.
A student looks at two distant point sources of light. The wavelength of each source is 590 nm. The angular separation
–4
between these two sources is 3.6 × 10 radians subtended at the eye. At the eye, images of the two sources are formed by
the eye on the retina.
(a) Estimate the diameter of the circular aperture of the eye.
(b) Use your estimate in (b) to determine whether the student can resolve these two sources. Explain your answer.
IB PHYSICS UNIT 11 RESOLUTION
SOLUTIONS
1. (a)
central maximum around X; correct overall shape (only two secondary maxima need be shown); secondary maxima
much less intense than the central maximum;
3
(b)
from d/D = 1/;2 = 10/13 = 500//;2
substitute to get λ = = 650 nm;
(c)
(i) the first minimum of one image should coincide with the first maximum of the other image; 1 max
(ii)
2.
(a)
2
use ; = /b to give b = 0.15 mm;
2
diffraction / bending of waves (due to an aperture / obstacle);
1
(b)
diffraction pattern falls on the maximum of the other.
Labels should include: maxima 1; maxima 2; minima 2 and minima 1
A note saying” just resolved” or similar
Note that a well-labelled and annotated diagram could receive up to 3 max
[4]
3.
(a)
(i) diffraction at the lens;
(ii)
circular patch – bright; circular bright ring/darkness between patch and ring;
2
(i) θ = ;
(b)
= 2.4 × 10
(ii)
4.
1
–4
rad;
2
–4
1.22 = 2.4 ×10 therefore d = ;
d = 2.8 mm;
Award [2] even if factor 1.22 is missing.
2
(a)
the diffraction pattern of one point source has its central maximum on the first minimum of the diffraction pattern of
the other point source;
2
Full marks can be awarded for a clearly drawn and fully labelled diagram. Partial credit is for answers that have some idea but
lack precision.
(b)
3 (± 2) mm;
1
(c)
correct calculation of Rayleigh criteria angle;
–9
–4
e.g. θ = 1.22 = 1.22 × 590 × 10 / 0.003 = 2.4 × 10 radians.
Accept answers that miss the factor of 1.2 to get 2.0 ×10
–4
radians.
correct comparison and answer;
2 max
e.g. this will be resolved as minimum angle is less than the separation of the point sources. Watch for ECF – this angle may or
may not be resolved depending on the estimation of the diameter of the aperture.