A beam of vertically polarised light is incident normally on a polarising filter. The filter can berotated so that it is always in a plane perpendicular to the beam. The transmission axis of thefilter is initially vertical.beam of verticallypolarised light wavestransmission axisof polarising filterpolarising filter transmitted lightThe filter is first rotated clockwise by an angle of 30° so that the transmitted light waves haveintensity I30 .The filter is then rotated clockwise by a further angle of 30°.What is the new intensity of the transmitted light waves?A 0.25 I30 B 0.33 I30 C 0.75 I30 D 0.87 I30
Question
A beam of vertically polarised light is incident normally on a polarising filter. The filter can berotated so that it is always in a plane perpendicular to the beam. The transmission axis of thefilter is initially vertical.beam of verticallypolarised light wavestransmission axisof polarising filterpolarising filter transmitted lightThe filter is first rotated clockwise by an angle of 30° so that the transmitted light waves haveintensity I30 .The filter is then rotated clockwise by a further angle of 30°.What is the new intensity of the transmitted light waves?A 0.25 I30 B 0.33 I30 C 0.75 I30 D 0.87 I30
Solution
The intensity of light transmitted through a polarizer can be calculated using Malus's law, which states that the intensity I of the transmitted light is proportional to the square of the cosine of the angle θ between the transmission axis of the polarizer and the plane of polarization of the incident light.
In this case, the light is initially vertically polarized and the transmission axis of the filter is also vertical, so the initial angle θ is 0° and the initial intensity I0 is the maximum possible intensity.
When the filter is rotated clockwise by 30°, the angle θ becomes 30° and the intensity I30 of the transmitted light is given by I30 = I0 cos²(30°).
When the filter is then rotated clockwise by a further 30°, the angle θ becomes 60° and the new intensity I60 of the transmitted light is given by I60 = I0 cos²(60°).
However, we want to find the new intensity relative to I30, not I0. To do this, we can divide the equation for I60 by the equation for I30 to get I60/I30 = cos²(60°) / cos²(30°).
Calculating this gives I60/I30 = (1/2) / (3/4) = 2/3 = 0.67.
Therefore, the new intensity of the transmitted light waves is 0.67 I30, which is not one of the given options. There may be a mistake in the question or the given options.
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