In experiment 1, unpolarized light falls on the polarizer in the figure below. T
ID: 2263249 • Letter: I
Question
In experiment 1, unpolarized light falls on the polarizer in the figure below. The angle of the analyzer is ? = 62.7
In experiment 1, unpolarized light falls on the polarizer in the figure below. The angle of the analyzer is ? = 62.7 degree . In experiment 2, the unpolarized light is replaced by light of the same intensity, but the light is polarized along the direction of the polarizer's transmission axis. By how many additional degrees must the analyzer be rotated so that the light falling on the photocell will have the same intensity as it did in experiment 1? Explain whether ? is increased or deceased by this additional number of degrees. (Use a positive number to indicate an increase in the angle. Use a negative number to indicate a decrease in the angle.)Explanation / Answer
Let the intensity of unpolarised light be Io. the intensity which cm out of teh first polriser can be assumed to be Io/2 out of this the inetnsity which crosses analyser - (Io/2)*cos^2 62.7.
Next when polarised light of same unetnsity falls on polariser, the same intensity I0o passes through. If now analyser makes an angle theta, te intensity that will cross through = Io*cos^ theta.
We have
Io*cos^2 theta = (Io/2)*cos^2 62.7 or
cos theta = [cos 62.7]/sqrt(2) = 0.3243 or theta = 71.07 degree. the anlyser must be rotated through
71.07 - 62.7 = 8.37 degree