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Light waves are electromagnetic waves and hence transverse in nature. The electric field in a light wave propagating in free space is perpendicular to the direction of propagation. But there are infinite number of directions perpendicular to the direction of propagation. but there are infinite number of directions perpendicular to the direction of propagation of light. for example if light propagates along x-axis, the electric field may be along y-axis or along the z-axis or along any direction in y-z plane in the ordinary light. However, if electric field vecE remains parallel to a fixed direction (say y-axis) then such light is called linearly polarised light. There are several methods to produce polarised light from the unpolarised light. now-a-days polaroid sheets are commonly used to produce linearly polarised light. a polaroid has long chain of hydrocarbons which become conducting at optical frequencies. when light falls normally on the polaroid sheet, the vecE parallel to the chains is absorbed in setting up electric currents in the chains but vecE perpendicular to the chain gets transmitted. so, light on passing through the polaroid i.e., the transmitted light become linearly polarised with vecE parallel to the transission (pass) axis of polaroid. If linearly polarised light of intensity 'I' is incident on another polaroid whose pass axis is inclined at an angle theta from pass axis of first polaroid (or transmission axis of vecE of linearly polarised light) the intensity of transmitted light I_(t) is given as: I_(t)=Icos^(2)theta. Q. Two polaroids P_(1) and P_(2) are set with their pass axis inclined at an angle 30^(@). if I_(0) be the light incident on P_(1), what is the intensity of light transmitted through P_(2) ? |
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Answer» Solution :`because` Intensity of LIGHT INCIDENT on polaroid `P_(1)=I_(0)` `therefore` Intensity of light after passing through `P_(1),I_(1)=(I_(0))/(2)` Since axis of polaroid `P_(2)` is inclined at `theta=30^(@)` from `P_(1)`, hence intensity of light being transmitted through `P_(2)` is: `I_(2)=I_(1)cos^(2)(30^(@))=(I_(0))/(2)xx((sqrt(3))/(2))^(2)=(3)/(8)I_(0)` |
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