(a) How does one demonstrate, using a suitable diagram, that unpolarised light when passed through a Polaroid gets polarised?
(b) A beam of unpolarised light is incident a glass-air interface. Show using a suitable ray diagram, that light reflected from the interface is totally polarised, when `mu=tani_(B)`, when `mu` is the refractive index of glass with respect to air and `i_(B)`, is the Brewster's angle.

Explanation

(a) (i) Two independent monochromatic sources of light cannot produce a sustained interference patter. The phase difference between these two sources will continuously vary, and the positions of maxima and minima will change with time.
(ii) `y_(1)=a cos omega t, and y_(2)=a cos (omegat+phi)`
`y=y_(1)+y_(2)=a [cos omega t+cos (omega t+phi)`
`=2acos((phi)/(2))cos(omegat+(phi)/(2))`
The resultant amplitude is `A=2a cos ((phi)/(2))`
and hence intensity `(I)=4a^(2)cos^(2)((phi)/(2))=4I_(0)cos^(2)((phi)/(2))`.
The intensity of light is directly proportional to the square of the amplitude of the wave. intensity of light at point P on the screen is given by :
`I=4a^(2) cos^(2)(phi)/(2)`.
(a)

The phenomenon of restricting the vibration of light (electric vector) in a particular direction perpendicular to the direction of the wave propagation is called polarisation of light.
When unpolarised light is passed through a polaroid, only those bibrations of light pass through the crystal, which are parallel to the axis of the crystal (AB). All other vibrations are absorbed and that is why intensity of the emerging light is reduced.
The plane ABCD in which the vibrations of the polarise light are confined is called the plane of vibration.
The plane KLMN that is perpendicular to the plane of vibration is defined as the plane of polarization. ,brgt (b) When unpolarised light is incident on the glass-air interface at Brewster angle iB, then reflected light is totally polaised. This is called Brewster's Law.
When light is incident at Brewster angle, the reflected component OB and the refracted component OC are mutually perpendicualr to each other.
From the figure, we have:
`angleBOY+angleYOC=90^(@)`
`(90^(@)-i_(B))+(90^(@)-r)=90^(@)`
where, r is angle of refraction
`90^(@)-i_(B)=r`
according to the Snell's law :
`mu=sin i-sin ri=i_(B) and`
`r=90^(@)-i_(B)mu=sini_(B) sin 90^(@)-i_(B)`
`sini_(B)cosi_(B)mu=tani_(B)`.
Hence, Proved.