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]
 
▶️Answer/Explanation

Markscheme

a.

diffraction angle=0.05 rad;
\(b = \left( {\frac{\lambda }{\theta } = \frac{{7.0 \times {{10}^{ – 7}}}}{{0.050}} = } \right)1.4 \times {10^{ – 5}}\left( {\rm{m}} \right)\);
(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.

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;

Scroll to Top