# IB DP Physics 9.2 – Single-slit diffraction Question Bank SL Paper 3

Question

This question is about diffraction and resolution.
Monochromatic light is incident on a narrow rectangular slit.

The light is observed on a screen far from the slit. The graph shows the variation with angle θ of the relative intensity for light of wavelength 7.0×10–7m.

a.

Estimate the width of the slit.

[2]
b.

On the graph, sketch the variation of the relative intensity with θ when the wavelength of the light is reduced.

[1]
c.

State and explain, with reference to your sketch in (b), whether it is easier to resolve two objects in blue light or in red light.

[2]

## Markscheme

a.

$$b = \left( {\frac{\lambda }{\theta } = \frac{{7.0 \times {{10}^{ – 7}}}}{{0.050}} = } \right)1.4 \times {10^{ – 5}}\left( {\rm{m}} \right)$$;
(do not accept use of 1.22)

Award [2] for a bald correct answer.

b.

same shape with narrower central maximum;
Ignore height of intensity peak.

c.

blue light gives better resolution;
blue light has shorter wavelength than red light;
giving smaller angle of diffraction;
Allow reverse argument for red light.

## Examiners report

Parts (a) and (b) on single slit diffraction were well answered.

b.

Parts (a) and (b) on single slit diffraction were well answered.

c.

However there were fewer correct answers for part (c) where effect of the different wavelengths of red and blue light were sometimes confused and the smaller θ interpreted as poorer resolution. Often the ability to resolve was explained incorrectly in terms of the intensity of the graphs drawn.

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]

## 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 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]

## 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.