Date | November 2011 | Marks available | 3 | Reference code | 11N.3.SL.TZ0.3 |
Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Estimate | Question number | 3 | Adapted from | N/A |
Question
This question is about diffraction and polarization.
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.
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.
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.
Markscheme
(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;