Discuss the intensity of transmitted light when a Polaroid sheet is rotated between two crossed polaroids ?

Q: Discuss the intensity of transmitted light when a Polaroid sheet is rotated between two crossed polaroids ?

Sol: let I_0 be the intensity of polarised light after passing through the first polariser P1 . Then the intensity P2 will be

I = I2 cos2 θ, Where θ is the angle between pass axes of P1 and P2 .

Since P1 and P2 are crossed the angle between the pass axes of P2 and P3 will be (π/2- θ). Hence the intensity of light emerging from P3 will be

I = I0 cos2 θ cos2 (π/2-θ)

= I0 cos2 θ sin2 θ = (I0/4) sin2 2 θ

Therefore, the transmitted intensity will be maximum when 𝜃 = π⁄4.

Unpolarized light of intensity 32 Wm^(-2) passes through three polarizers such that the transmission axis of the last polarizer is crossed with the first. If the intensity of the emerging light is 3 Wm^(-2), what is the angle between the transmission axes…

Q: Unpolarized light of intensity 32 Wm-2 passes through three polarizers such that the transmission axis of the last polarizer is crossed with the first. If the intensity of the emerging light is 3 Wm-2, what is the angle between the transmission axes of the first two polarizers ? At what angle will the transmitted intensity be maximum?

Sol: If θ is the angle between the transmission axes of first Polaroid P1 and second P2 while φ between the
transmission axes of second Polaroid P2 and third P3 , then according to give problem.

φ = (90° – θ) … (i)

Now id I_0 is the intensity of Unpolarized light incident on Polaroid P_1, the intensity of light transmitted through it,

I1 = I0/2 = (32)/2 = 16 W/m2 … (ii)

Now as angle between transmission axes of polaroids P1 and P2 is θ , in a accordance with Malus law, intensity of light transmitted through P2 will be

I2 = I1 cos2 θ = 16 cos2 θ …. (iii)

And as angle between transmission axes of P2 and P3 is φ , light transmitted through P3 will be

I3 = I2 cos^2 φ = 16 cos2 θ cos2 φ … (iv)

According to give problem, I3 = 3 W/ m2

So, 4(sin 2θ)2 = 3

i.e., sin 2 θ = (√3/2)

or, 2 θ = 60°, i.e., θ = 30°.

Unpolarized light falls on two sheets placed one on top of the other. What must be the angle between the characteristic direction of the sheets if the intensity of the transmitted light is one third of intensity of the incident beam ?

Q: Unpolarized light falls on two sheets placed one on top of the other. What must be the angle between the characteristic direction of the sheets if the intensity of the transmitted light is one third of intensity of the incident beam ?

Sol: Intensity of the light transmitted through the first polarized I1 = I0/2 , where I0 is the intensity of the incident unpolarized light. Intensity of the light transmitted through the second polarize is I2 = I1 cos2θ

θ is the angle between the characteristic directions of the polarizer sheets.

But I2 = I0/ 3 (given)

∴ I2 = I1 cos2θ = (I0/2) cos2 θ

I0/ 3 = (I0/2) cos2 θ

∴ cos2 θ = 2/3

⇒ $\large \theta = cos^{-1}\sqrt{\frac{2}{3}} $

When light of a creation wavelength is incident an a plane surface of a material at a glancing angle 30°, the reflected light is found to be completely plane polarized determine…

Q: When light of a creation wavelength is incident an a plane surface of a material at a glancing angle 30°, the reflected light is found to be completely plane polarized determine.
(a) refractive index of given material and
(b) angle of refraction.

Sol: (a) Angle of incident light with the surface is 30°. The angle of incidence = 90° – 30° = 60°. Since
reflected light is completely polarized, therefore incidence takes place at polarizing angle of incidence
ip.

∴ ip = 60°

Using Brewster’s law ,

𝜇 = tan ip

𝜇 = tan 60° = √3

(b) From Shell’s law

$\large \mu = \frac{sini}{sinr} $

$\large \sqrt{3} = \frac{sin60^o}{sinr} $

$\large sinr = \frac{sin60^o}{\sqrt{3}} = \frac{\sqrt{3}}{2 \times \sqrt{3}} $

or sin r = 1/2,

r = 30°

Two slits are made one millimetre apart and the screen is placed one meter away. What should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern.

Q: Two slits are made one millimetre apart and the screen is placed one meter away. What should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern.

Sol: $\large 10 \frac{\lambda}{d} = 2 \frac{\lambda}{a} $

$\large a = \frac{d}{5} = \frac{1}{5} $

∴ a = 0.2 mm