After the class is over you take a journey to a tropical oasis to get some rest
ID: 1998757 • Letter: A
Question
After the class is over you take a journey to a tropical oasis to get some rest
and relaxation. While staring at the sunset, you notice that the reflection of
the sun off the surface of the water (which you can approximate as flat) is
slowly disappearing from your view. Then you remember, your totally fly
and not at all hastily drawn on sunglasses are linearly polarized. So…
a.) What intensity of direct sunlight (not reflected) passes through
your glasses? Assume direct sunlight has an intensity of ID.
b.)For your glasses to completely cut the reflection of the sun off the
surface of the water, how must their axis of polarization be oriented relative
to the surface of the water? Circle one:
Perpendicular // 45° // Horizontal
c.) At what angle above the horizon will the sun be when the
reflection totally disappears from your sight, i.e. will it be totally linearly
polarized and blocked by properly made glasses? Air has n = 1.00, water has
n = 1.33.
d.)If your glasses are tilted at an angle of 60° relative to the
polarization of the reflected light, what intensity will pass through? Assume
the reflected/polarized sunlight off the water has intensity IR.
Explanation / Answer
(a) Direct Sunlight(unpolarised) will pass through Sunglasses with an Internsity (ID/2). where ID is internsity of Sunlight.
(b) When unpolarised light incidents a flat surface at some particular angle(Brewster) it becomes polarised.
Well if sunlght after reflection through water surface enters sunglasses,to block this sunlight,its axis of polarization should be perpendicular to that of water surface.
(c) Now we calculate Brewster Angle :
Tan x = nwater /nair
x is brewster angle , so x is Tan-1 (nw/nair) =53.06 degrees.
(d) This is given by Malus law :
So, I60* = Irefl. through water Cos2(60)
I60* = IR * 0.52 = 0.25*IR.