IB DP Physics 9.4 – Resolution Question Bank SL Paper 3

 

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Question

This question is about diffraction and resolution.

Two identical sources of electromagnetic radiation, S1 and S2, emit monochromatic coherent waves of wavelength 59 μm. The waves pass through a circular aperture and are incident on a screen.

S1 and S2 are at the same distance from the aperture. The diameter of the aperture is 0.18 mm. The angle between the lines joining the sources to the aperture is 0.25 rad.

S1 is turned on and S2 is turned off.

(i) Show that the angle at which the first minimum of the diffraction pattern occurs is 0.40 rad.

(ii) On the axes below, sketch a graph to show how the intensity I of the radiation from S1 varies with the diffraction angle θ.

Answer/Explanation

Markscheme

(i) \(\theta  = \frac{{1.22 \times 59 \times {{10}^{ – 6}}}}{{0.18 \times {{10}^{ – 3}}}}\);
(=0.40rad)

(ii) central symmetrical maximum;
at least one secondary maximum on each side, no more than one third the height of the central maximum; (judge by eye)
minima drawn to zero, ie touching axis at θ=±0.40 (and ±0.80);

 
Question

This question is about standing waves and the Doppler effect.

The horn of a train can be modeled as a pipe with one open end and one closed end. The speed of sound in air is 330ms–1.

a.

On leaving the station, the train blows its horn. Both the first harmonic and the next highest harmonic are produced by the horn. The difference in frequency between the harmonics emitted by the horn is measured as 820 Hz.

(i) Deduce that the length of the horn is about 0.20 m.

(ii) Show that the frequency of the first harmonic is about 410 Hz.

[5]
b.

(i) Describe what is meant by the Doppler effect.

(ii) The train approaches a stationary observer at a constant velocity of 50ms–1 and sounds its horn at the same frequency as in (a)(ii). Calculate the frequency of the sound as measured by the observer.

[4]
 
Answer/Explanation

Markscheme

a.

(i) \({f_1} = \frac{v}{{4L}}\), \({f_2} = 3{f_1} = \frac{{3v}}{{4L}}\);
\({f_2} – {f_1} = \frac{v}{{2L}} = 820\left( {{\rm{Hz}}} \right)\);
\(L = \frac{{330}}{{2 \times 820}}\);
(L=0.20m)

(ii) \(\lambda = 4L = 0.80\left( {\rm{m}} \right)\);
\(f = \left( {\frac{{330}}{{0.8}}} \right) = 413 {\rm{Hz}}\);

This is a question testing units for this option. Do not award second marking point for an incorrect or missing unit.

b.

(i) a change in the observed frequency/wavelength of a wave;
when there is relative motion of observer and source;

(ii) \(f’\left( { = f\frac{v}{{v – {u_s}}}} \right) = 410 \times \frac{{330}}{{330 – 50}}\);
\(f’ = 480\left( {{\rm{Hz}}} \right)\);
Allow ECF from (a)(ii).

Question

This question is about diffraction and resolution.

Monochromatic light is incident normally on a single narrow slit and gives rise to a diffraction pattern on a screen.

a.

Sketch, for the diffraction pattern produced, a graph showing the variation of the relative intensity of the light with the angle measured from the centre of the slit.

[2]
b.

The single narrow slit is replaced by a double narrow slit. Explain, with reference to your answer to (a), how the Rayleigh criterion applies to the diffraction patterns produced by the light emerging from the two slits.

[3]
c.

Two lamps emit light of wavelength 620 nm. The lights are observed through a circular aperture of diameter 1.5 mm from a distance of 850 m. Calculate the minimum distance between the lamps so that they are resolved.

[2]
 
Answer/Explanation

Markscheme

a.

large central peak and at least one subsidiary maximum on each side;

minima have intensity of zero and intensity of secondary maxima at most 25 % of central maximum; } (judge by eye)

b.

explanation of resolving – seeing images as being from separate objects;

idea of diffraction patterns overlapping;

central maximum being at least as far from companion as the first minimum;

Marking points may be seen on graph in (a).

Marking points may be seen from diffraction pattern showing resultant intensity from two sources with a slight dip in the centre.

c.

equating \(1.22\frac{\lambda }{b}\) to \(\frac{x}{D}\);

0.43 (m);

Examiners report

a.

The majority could correctly sketch the diffraction pattern and only a few showed non-zero intensity at the minima. The definition of Rayleigh’s criterion was well known but candidates found it difficult to gain full marks explaining how the Rayleigh criterion applies to diffraction patterns, as asked. The calculation was well-attempted.

b.

The majority could correctly sketch the diffraction pattern and only a few showed non-zero intensity at the minima. The definition of Rayleigh’s criterion was well known but candidates found it difficult to gain full marks explaining how the Rayleigh criterion applies to diffraction patterns, as asked. The calculation was well-attempted.

c.

The majority could correctly sketch the diffraction pattern and only a few showed non-zero intensity at the minima. The definition of Rayleigh’s criterion was well known but candidates found it difficult to gain full marks explaining how the Rayleigh criterion applies to diffraction patterns, as asked. The calculation was well-attempted.

Question

This question is about resolution and polarization.

a.

A ship sails towards two stone towers built on land.

Emlyn, who is on the ship, views the towers. The pupils of Emlyn’s eyes are each of diameter 2.0 mm. The average wavelength of the sunlight is 550 nm.

State the Rayleigh criterion.

[2]
b.

(i) Calculate the angular separation of the two towers when the images of the towers are just resolved by Emlyn.

(ii) Emlyn can just resolve the images of the two towers when his distance from the towers is 11 km. Determine the distance between the two towers.

[3]
d.

Emlyn puts on a pair of polarizing sunglasses. Explain how these sunglasses reduce the intensity of the light, reflected from the sea, that enters Emlyn’s eyes.

[2]
 
Answer/Explanation

Markscheme

a.

for the images (of two sources) just to be resolved/distinguished/seen as separate;

central maximum of one diffraction pattern must coincide with first minimum of second / OWTTE;

Accept a suitably drawn diagram for the second marking point.

b.

(i) \(\theta = \left( {\frac{{{\text{1.22}} \times {\text{550}} \times {\text{1}}{{\text{0}}^{ – 9}}}}{{{\text{2.0}} \times {\text{1}}{{\text{0}}^{ – 3}}}} = } \right){\text{ }}3.4 \times {10^{ – 4}}{\text{ (rad)}}\)\(\,\,\,\)or\(\,\,\,\)0.019°;

(ii) \(d = 11 \times {10^3} \times 3.4 \times {10^{ – 4}}\);

\( = {\text{3.7 (m)}}\);

Award [2] for a bald correct answer.

d.

reflected light is (partially) polarized parallel to sea surface/horizontally polarized;

sunglasses have a transmission axis at 90° to reflected light/vertical transmission axis;

Examiners report

a.

An improvement in the answers to (a) was noted.

b.

In (b) very few omitted to use the factor of 1.22 and full marks were often scored for both calculations.

d.

(d) was poorly answered with many being unable to make it clear that reflected light from the sea would be partially horizontally polarized. Some just referred to the darkness of the sunglasses’ lens.

 
Question

This question is about radio telescopes.

A distant galaxy emits radio waves of frequency 6.0×109 Hz and is moving with speed 6.0×106 ms–1 directly away from an observer on Earth.

a.

Determine the wavelength of the radio wave as measured by the observer on Earth.

[3]
b.

The radio signals from two stars on opposite sides of the galaxy are detected on Earth using a radio telescope. The telescope has a circular receiving dish.

(i) State the Rayleigh criterion for the images of two point sources to be just resolved.

(ii) The galaxy is 2.0×1021m from Earth and the stars are separated by 5.0×1019m. Determine the minimum size of the telescope dish required to resolve the images of the two stars at a wavelength of 5.1×10–2m.

[4]
Answer/Explanation

Markscheme

a.

\(\left( {{\rm{since}}\frac{v}{c} \ll 1} \right)\)\(\Delta f\left( { = \frac{v}{c}f} \right) = \frac{{6.0 \times {{10}^6}}}{{3.0 \times {{10}^8}}}6.0 \times {10^9}\left( { = 0.12 \times {{10}^9}{\rm{Hz}}} \right)\);
\(f’ = f – \Delta f\left( { = 5.9 \times {{10}^9}{\rm{Hz}}} \right)\);
\(\lambda ‘\left( { = \frac{c}{{f’}} = \frac{{3.00 \times {{10}^8}}}{{5.9 \times {{10}^9}}}} \right) = 5.1 \times {10^{ – 2}}\left( {\rm{m}} \right)\);

Award [2 max] if ƒ’=ƒ+Δƒ used to give λ=4.9×10-2(m).

b.

(i) the two (point-like) sources generate diffraction patterns with central maxima;
the central maximum of one pattern overlaps with the first minimum of the second diffraction pattern;

(ii) \(\theta \approx \frac{d}{D} = \frac{{5.0 \times {{10}^{19}}}}{{2.0 \times {{10}^{21}}}} = 0.025({\rm{rad}})\)
\(\left( {b > 1.22\frac{{5.1 \times {{10}^{ – 2}}}}{{0.025}} = } \right)2.5\left( {\rm{m}} \right)\);

Allow [1 max] for solution that omits 1.22.

Examiners report

a.

was well done by some, although many used the equation appropriate for sound not the approximation for light where c >> v.

b.

In (b) (i) candidates often got the first mark by implication when gaining second mark. There was a lot of confused algebra in (b)(ii).

Question

This question is about resolution.

Light from two monochromatic point sources passes through a circular aperture and is observed on a screen.

The graph shows how the intensity I of the light on the screen varies with the angle θ .

The two sources are just resolved according to the Rayleigh criterion.

a.

State what is meant by resolved in this context.

[1]
b.

The wavelength of the light from the two sources is 528 nm. The distance of the two sources from the aperture is 1.60 m.

Using data from the graph opposite, determine the

(i) separation of the two sources.

(ii) diameter of the aperture.

[3]
 
Answer/Explanation

Markscheme

a.

the two sources are seen as two distinct sources / two distinct images are formed / the central maximum of one source coincides with the first minimum of the other;

b.

(i) realization that the diffraction angle of the one source diffraction pattern is at 0.008 radians;
and so separation (=1.60×0.008=1.28)≈0.013m;

(ii) \(\left( {1.22\frac{{528 \times {{10}^{ – 9}}}}{b} = 0.008 \Rightarrow } \right)b = 0.081{\rm{mm}}\);

Question

This question is about diffraction and polarization.

a.

Light from a monochromatic point source S1 is incident on a narrow, rectangular slit.

After passing through the slit the light is incident on a screen. The distance between the slit and screen is very large compared with the width of the slit.

(i) On the axes below, sketch the variation with angle of diffraction θ of the relative intensity I of the light diffracted at the slit.

(ii) The wavelength of the light is 480 nm. The slit width is 0.1 mm and its distance from the screen is 1.2 m. Determine the width of the central diffraction maximum observed on the screen.

[5]
b.

Judy looks at two point sources identical to the source S1 in (a). The distance between the sources is 8.0 mm and Judy’s eye is at a distance d from the sources.

Estimate the value of d for which the images of the two sources formed on the retina of Judy’s eye are just resolved.

[3]
c.

The light from a point source is unpolarized. The light can be polarized by passing it through a polarizer.

Explain, with reference to the electric (field) vector of unpolarized light and polarized light, the term polarizer.

[3]
 
Answer/Explanation

Markscheme

a.

(i)

overall correct shape with central maxima at θ=0; { (only one secondary maximum required each side of θ=0)
secondary maximum no greater than ¼ intensity of central maximum; { (judge by eye)

(ii) \(\theta = \frac{\lambda }{b} = \frac{x}{D}\) (where x is the half width of central maximum);
\(2x = 2\frac{{D\lambda }}{b}\);
\(\left( {\frac{{2 \times 1.2 \times 4.8 \times {{10}^{ – 7}}}}{{{{10}^{ – 4}}}}} \right) = 12{\rm{mm}}\);

b.

diameter of pupil =3.0 mm; (accept answers in the range of 2.0 mm to 5.0 mm)
\(\theta = \left( {1.22 \times \frac{\lambda }{b} = 1.22 \times \frac{{4.8 \times {{10}^{ – 7}}}}{{3.0 \times {{10}^{ – 3}}}} = } \right)1.95 \times {10^{ – 4}}\left( {{\rm{rad}}} \right)\);
\(d = \frac{{8.0 \times {{10}^{ – 3}}}}{{1.95 \times {{10}^{ – 4}}}} = 41{\rm{m}}\); (accept answer in the range of 20m to 70m)

c.

in unpolarized light the plane of vibration of the electric (field) vector is continually changing / OWTTE;
in polarized light the electric vector vibrates in one plane only;
a polarizer is made of material that absorbs/transmits either the horizontal or vertical component/only one component of the electric vector;

Question

This question is about resolution.

A car is travelling along a straight road at night. To a distant observer the two headlamps of the car appear as a single point source. With the aid of an appropriately labelled sketch graph, explain this observation.

Answer/Explanation

Markscheme

graph: [2 max]
correct single slit diffraction pattern with appropriate labelling (eg angle/θ) of horizontal axis;
identical diffraction pattern with peak separated by less than half the width of central maximum;

explanation: [2 max]
light from each lamp produces a diffraction pattern on retina of eye;
some statement/Rayleigh criterion quoted to the effect that because the minimum of one falls too close to the maximum of the other they cannot be resolved, so diffraction pattern looks the same as that from a single lamp/source;

Question

This question is about resolution.

a.

Two point sources S1 and S2 emit monochromatic light of the same wavelength. The light is incident on a small aperture A and is then brought to focus on a screen.

The images of the two sources on the screen are just resolved according to the Rayleigh criterion. Sketch, using the axes below, how the relative intensity I of light on the screen varies with distance along the screen d.

[3]
b.

A car is travelling at night along a straight road. Diane is walking towards the car. She sees the headlights of the car as one single light. Estimate, using the data below, the separation d between Diane and the car at which, according to the Rayleigh criterion, Diane will just be able to see the headlights as two separate sources.

Distance between the headlights = 1.4 m
Average wavelength of light emitted by the headlights = 500 nm
Diameter of the pupils of Diane’s eyes = 1.9 mm

[3]
c.

The light from the car headlights in (b) is not polarized. State what is meant by polarized light.

[1]
 
Answer/Explanation

Markscheme

a.

correct shape of two diffraction patterns showing central maximum and at least one secondary maximum each side of central maximum;
intensity of secondary maxima no greater than one third intensity of central maxima; } (judge by eye)
first minimum of one pattern coincident with central maximum of other pattern;

or

Allow just the approximate dotted resultant intensity patterns:
correct pattern of two symmetrical principal maxima;
with local minimum between them;
at least one secondary maximum on each side which are no more than \(\frac{1}{3}\) of the intensity of the principal maxima;

b.

angular separation for resolution=\(1.22\frac{\lambda }{b} = 1.22 \times \frac{{5.0 \times {{10}^{ – 7}}}}{{1.9 \times {{10}^{ – 3}}}} = \left( {3.21 \times {{10}^{ – 4}}} \right)\left( {{\rm{rad}}} \right)\);
=\(\frac{{1.4}}{d}\);
d=4.4(km);
Award [2 max] if 1.22 not used and answer is 5.3 km.
Award [3] for a bald correct answer.

c.

light in which the electric/magnetic field (vector) vibrates only in one plane/direction;

Question

This question is about diffraction and resolution.

A parallel beam of monochromatic light is incident on a narrow rectangular slit. After passing through the slit, the light is incident on a distant screen.

Point X is the midpoint of the slit.

a.

(i) On the axes below, sketch a graph to show how the intensity of the light on the screen varies with the angle \(\theta \) shown in the diagram.

(ii) The wavelength of the light is 520 nm, the width of the slit is 0.04 mm and the screen is 1.2 m from the slit. Show that the width of the central maximum of intensity on the screen is about 3 cm.

[5]
b.

Points P and Q are on the circumference of a planet as shown.

By considering the two points, outline why diffraction limits the ability of an astronomical telescope to resolve the image of the planet as a disc.

[3]
 
Answer/Explanation

Markscheme

a.

(i)

general correct shape touching axis and symmetric about \(\theta = 0\) (at least onesecondary maxima on each side); (judge by eye)

central maximum wider than secondary maxima;

secondary maxima at most one third intensity of central maximum;

(ii) \(\frac{d}{2} = \frac{{D\lambda }}{b}\);

\(d = \frac{{2.0 \times 1.2 \times 5.2 \times {{10}^{ – 7}}}}{{4.0 \times {{10}^{ – 5}}}} = 3.12 \times {10^{ – 2}}{\text{ m}}\);

\( \approx 3{\text{ cm}}\)

b.

Award [2 max] for a sensible argument.

e.g. light from each point forms a diffraction pattern after being focussed by the eyepiece of the telescope;

if the diffraction patterns are not sufficiently well separated then the points will not be resolved as separate sources;

Award [1 max] for the conclusion.

e.g. if the points cannot be resolved as separate sources the planet cannot be seen as a disc;

Examiners report

a.

Intensity distributions were often drawn well but quite a few candidates did not have their graphs in contact with the \(\theta \) axis. Candidates’ working was often difficult to follow in the calculation part of this question.

b.

Very few candidates recognised the role played by diffraction in the resolution of the planet as a disc.