Suppose you have a polarized light source in your lab that has an intensity of 1

Question

Suppose you have a polarized light source in your lab that has an intensity of 1916 W/m2, where the electric field oscillates in the z direction. The light travels toward a workbench and encounters two polarizing filters.

(a) If the light passes first through a filter with its transmission axis at an angle of 45.0° from the z direction, and then through a filter with its transmission axis at an angle of 90.0° from the z direction, what is the final intensity of the transmitted light?
W/m2

(b) If, instead, the light passes first through a filter with its transmission axis at an angle of 90.0° from the z direction, and then through a filter with its transmission axis at an angle of 45.0° from the z direction, what is the final intensity of the transmitted light?
W/m2

Explanation / Answer

Malus’s law. Let the angle between the transmission axis of the polarizer and that of the analyzer be . Then the amount of light I transmitted through the analyzer depends on the value of as follows:

1)        I = Imax cos2

where Imax is the maximum amount of light transmitted i.e. the amount transmitted at = 0. This law was discovered experimentally by Etienne Louis Malus in 1809.

part a

for crystal 1

I1=1916cos245=958wb/m2

for crystal 2

now polarized light is already at angle of 45 from Z axis so angle between the crystal 2 and direction of movin is 45 degree

Ifinal=I1cos245=479Wb/m2

part b

prcictacally when unpolrised light passes through perpendicular crystal intensity become half and direction is parallel to axis of crystal

So I1=I/2=958Wb/m2

Ifinal=958cos245=479Wb/m2

only differences on both part is direction of polarised light and intensity is same

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