Saturday, March 03, 2007

Questions on Polarisation -Brewster’s law

Most of you might have noted that the transverse nature of light wave was proved by the phenomenon of polarisation. You should remember that sound wave cannot be polarised since it is longitudinal.
You will often find questions based on Brewster’s law in the section on polarisation. When unpolarised light proceeding through a rarer medium is incident at an angle ‘i’ on a denser medium, the reflected beam is plane polarised if tan i = n where ‘n’ is the refractive index of the denser medium with respect to the rarer medium.. This is Brewster’s law. [ The transmitted beam will be partially plane polarised].
Here is a simple question based on Brewster’s law:
When unpolarised beam of light is incident on a glass slab, the rflected beam is found to be completely plane polarised. The angle between the reflected beam and the transmitted beam is
(a) 45° (b) 45° (c) 60° (d) 90° (e) dependent on the refractive index
The correct option is (c). You can easily prove this as follows:
Since n = sin i/sin r, we have tan i = sin i/ sin r so that sin i/ cos i = sin i/sin r. Therefore, cos i = sin r. Therefore, r = 90° – i so that i + r = 90°. With reference to the figure, the angle BOC ( which is the angle between the reflected and transmitted beams) is therefore equal to 90°.
Now consider the following question:
When unpolarised light proceeding through air is incident at an angle of 60° on a transparent slab, the reflected rays are found to be completely plane polarised. The refractive index of the slab and the angle of refraction into the slab are respectively
(a) 1.7, 30° (b) 1.414, 40° (c) 1.5, 30° (d) 1.3, 45° (e) 1.732, 30°
If you are in too much hurry, you may pick out option (a). But the correct option is (e). Refractive index, n = tan i = tan 60° = √3 = 1.732. Angle of refraction r = 90° – i = 90° – 60° = 30°.

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