We know that in case of unpolarised light, `vecE` vectors (light vectors) oscillate periodically with equal probability in all the directions in a plane perpendicular to direction of propagation of light. In the figure, only one `vecE` vector is shown as a sample, which oscillates in YZ plane, shown at a distance x from light source placed at origin. Its displacement at time t is suppose `vecE`.
Above vector can be decomposed into two mutually perpendicular components `vece_(n)` (perpendicular to Y-axis) and `vece_(t)` (parallel to Y-axis). Thus,
`vece=vece_(n)+vece_(t)" "....(1)`
Since `vece_(n)botvece_(t)`, we can write,
`e^(2)=e_(n)^(2)+e_(t)^(2)" "....(2)`
When two rays 1 and 2 superpose at some point on the screen, their normal components are mutually parallel and their tangential components are mutually parallel and so according to principle of superposition,
`e_(n)=(e_(n))_(1)+(e_(n))_(2)" "....(3)`
and `e_(t)=(e_(t))_(1)+(e_(t))_(2)" "....(4)`
n the presence of polaroid, normal component in ray 2 gets absorbed and so in the transmitted ray, `(e_(n))_(2)=0` and so from equation (3),
`e_(n)=(e_(n))_(1)" ....(5)`
Here, amplitudes of all `vecE` vectors oscillating in all the directions are equal. If this common equal amplitude is `E_(0)` then according to wave equation, for normal component in ray 1,
`(e_(n))_(1)=E_(0)sin(omegat-kx)" "......(6)`
For tangential component in ray 1,
`(e_(t))_(1)E_(0)sin(omegat-kx)" "....(7)`
For tangential component in ray 2,
`(e_(t))_(2)=E_(0)sin(omegat-kx+phi)" "......(8)`
(Where `phi=` phase difference between ray (1) and (2) at point P, at time t).
From equations (5) and (6),
`e_(n)=E_(0)sin(omegat-kx)" "......(9)`
`:.e_(n)^(2)=E_(0)^(2)sin^(2)(omegat-kx)" "....(10)`
From equations (4), (7), (8),
`:.e_(t)=E_(0)sin(omegat-kx)+E_(0)sin(omegat+kx-phi)` `:.e_(t)=E_(0)sin(omegat-kx)+E_(0){sin(omegat-kx)cosphi+cos(omegat-kx)sinphi}`
`:.e_(t)=E_(0)sin(omegat-kx)(1+cosphi)+E_(0)cos(omegat-kx)sinphi`
`:.e_(t)^(2)=E_(0)^(2)sin^(2)(omegat-kx)(1+cosphi)^(2)+E_(0)^(2)cos^(2)(omegat-kx)`
`sin^(2)phi+2E_(0)^(2)sin(omegat-kx)(1+cosphi)cos(omegat-kx)sinphi" "......(11)`
Placing the values from equation (10) and (11) into equation (2),
`e^(2)=E_(0)^(2)sin^(2)(omegat-kx){1+(1+cosphi)^(2)}+E_(0)^(2)cos^(2)(omegat-kx)sin^(2)phi+2E_(0)^(2)sin(omegat-kx)`
`(1-cosphi)cos(omegat-kx)sinphi`
`e^(2)=E_(0)^(2)sin^(2)(omegat-kx){2+2cosphi+cos^(2)phi)}+E_(0)^(2)cos^(2)(omegat-kx)sin^(2)phi+2E_(0)^(2)sin(omegat-kx)`
`(1+cosphi)cos(omegat-kx)sinphi`
Here in above equation first term, involving square of periodic function `sin(omegat-kx)` in terms of time is `E_(0)^(2)sin^(2)(omegat-kx)(2+2cosphi+cos^(2)phi)`
which gives us square of resultant amplitude of light vector at the point of superposition shown by `E_(R)^(2)`. Hence,
`E_(R)^(2)=E_(0)^(2)sin^(2)(omegat-kx)(2+2cosphi+cos^(2)phi)`
`:.ltE_(R)^(2)gt"= "E_(0)^(2)((1)/2)(2+2cosphi+cos^(2)phi)`
( `:.E_(0)andcosphi` is constant)
`:.ltE_(R)^(2)gt=(E_(0)^(2))/(2)(2+cosphi+cos^(2)phi)`
Now, if resultant intensity of light at the point of superposition is `I_(R)` then,
`I_(R)propltE_(R)^(2)gt`
`:.I_(R)=KltE_(R)^(2)gt`
`:.I_(R)=(KE_(0)^(2))/(2)(2+2cosphi+cos^(2)phi)" "....(12)`
Now, in the absence of polaroid, at the central point on the screen, resultant intensity is given lo in the statement, which is 4 times the intensity of one light ray. Hence,
`I_(0)=4IimpliesI=(I_(0))/(4)=KE_(0)^(2)" ".......(13)`
`("":.IpropE_(0)^(2))`
From equations (12) and (13),
`I_(R)=(I_(0))/(8)(2+2cosphi+cos^(2)phi)" ".....(14)`
Now, in the presence of polaroid, if intensity of light at the central point on the screen is `I_(max)` then to find its value, we should take `cosphi=max=1` in above equation.
Hence `I_(max)=(I_(0))/(8)(2+2+1)=(5I_(0))/(8)" ".....(15)`
Now, similarly in the presence of polaroid, if intensity of light at the first order minimum is `I_(min)` then to find its value, we should take `cosphi=min=-1` in eqaution (14). Hence,
`I_(min)=(I_(0))/(8)(2-2+1)=(I_(0))/(8)" "......(16)`
Equations (15) and (16) give required results.
