Wave Optics

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.

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°.

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}} $

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
i_p.

∴ i_p = 60°

Using Brewster’s law ,

𝜇 = tan i_p

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°

Q: Assume that light of wavelength 6000 Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?

Sol: objective has a diameter of 100 inch ,

a = 100 inch = 254 cm , λ ≈ 6000 A° = 6 × 10^-5 cm

∆θ = 1.22λ/a ≈ 2.9 × 10^-7 radians

Q: For what distance is ray optics α good approximation when the aperture is 3 mm wide and the wavelength is 500 nm ?

Sol: For distance Z ≤ Z_F, Ray optics is the good approximate ,

Fresnel distance , $\large Z_F = \frac{a^2}{\lambda} $

$\large Z_F = \frac{(3 \times !0^{-2})^2}{5 \times 10^{-7}} $

Z_F = 18 m

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