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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]
▶️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.
same shape with narrower central maximum;
Ignore height of intensity peak.
blue light gives better resolution;
blue light has shorter wavelength than red light;
giving smaller angle of diffraction;
Allow reverse argument for red light.
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]
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]
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);
\(2x = 2\frac{{D\lambda }}{b}\);
\(\left( {\frac{{2 \times 1.2 \times 4.8 \times {{10}^{ – 7}}}}{{{{10}^{ – 4}}}}} \right) = 12{\rm{mm}}\);
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)
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;
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