Contents
Physics Topics are often described using mathematical equations, making them precise and quantifiable.
What is Polaroid? What are the Uses of Polaroid?
Polarisation of light by dichroism: Some double refracting crystals like tourmaline have the property of converting ordinary unpolarised light into plane polarised light. This type of crystal is called dichroic crystal and this property of the crystal is known as dichroism.
When ordinary unpolarised light passes through a thin slice of a tourmaline crystal, it splits into two polarised rays—O-ray and E-ray. The plane of vibration of these two rays are perpendicular to each other. While passing through the crystal, O-ray is completely absorbed, but E-ray is transmitted from the crystal. The more is the thickness of the crystal, the more will be the absorption of O-ray and we get only polarised E-ray. This property of the crystal is called dichroism. So the emergent light is plane polarised.
Polaroid: Polaroid is a synthetic dichroic material. It is a polarising sheet or film by which polarised light can be produced. Polaroid films are produced by spreading ultramicroscopic crystals of herapathite on a thin sheet of nitrocellulose. The crystals are placed on the film by a special device in such a way that the optic axes of all th crystals are parallel. These crystals are highly dichroic and absorb one of the double refracted beams completely. The other refracted beam is transmitted from the crystals. So the transmitted light is plane polarised.
1. Herapathite is an organic compound whose chemical name is iodoquinine sulphate. It is a dichroic crystal used to polarise light rays.
2. Each polaroid has a particular plane of polarisation. A polaroid allows only those incident unpolarised light rays to refract, whose vibration plane is parallel to the polarisation plane of the polaroid. The direction of this polarisation plane of the polaroid is called transmission axis [Fig.].
Uses of polaroid:
- With the help of polaroids, plane polarised light can be produced and analysed very easily at a low cost.
- Polaroids are widely used as polarising sunglasses.
- Polaroids are used to eliminate the head light glare in motor cars.
- Polaroids are used as glass windows in trains and aeroplanes to control the Intensity of light coming from outside.
- In calculators and watches, letters and numbers are formed by liquid crystal display (LCD) through polarisation of light.
- Polaroid films are used in making 3D cinema or picture.
Numerical Examples
Example 1.
For producing a Fraunhofer diffraction fringe, a screen is placed 2 m away from a single narrow slit. If the width of slit is 0.2 mm, it is found that first minimum lies 5 mm on either side of the central maximum. Find the wavelength of the incident light.
Solution:
We know, if the distance between n th minimum and central maximum be xn, then
xn = \(\frac{n D \lambda}{a}\)
or, 0.5 = \(\frac{1 \times 200 \times \lambda}{0.02}\) [here n = 1, xn = 5 mm = 0.5 cm, D = 2 m = 200 cm, a = 0.2 mm = 0.02 cm]
∴ λ = \(\frac{0.5 \times 0.02}{1 \times 200}\) = 5000 × 10-8cm = 5000Å
Example 2.
A single narrow slit of width 0.1 mm, is illuminated with a parallel beam of light of wavelength 600 × 10-9m. An interference fringe is formed on a screen 40 cm away from the silt. At what distance, will the third minimum band be formed from the central maximum band?
Solution:
We know, if the distance of n th minimum from the central maximum be xn, then
xn = \(\frac{n D \lambda}{a}\) [here, n = 3, a = 0.1 mm = 0.01 cm, D = 40 cm and λ = 600 × 10-9 m = 600 × 10-7cm]
∴ x3 = \(\frac{3 \times 40 \times 600 \times 10^{-7}}{0.01}\) = 12 × 6 × 10-2 = 0.72 cm
Example 3.
A Fraunhofer diffraction pattern is formed by a light, of wavelength 600 nm, through a silt of width 1.2 µm. Find the angular position of the first minimum and angular width of the central maximum.
Solution:
If angular position of first minimum with respect to central maximum be θ, then
sinθ = \(\frac{\lambda}{a}\) = \(\frac{600 \times 10^{-9}}{1.2 \times 10^{-6}}\) = 0.5 = \(\frac{1}{2}\) = sin 30°
∴ θ = 30° (here, λ = 600 nm = 600 × 10-9m, a = 1.2 µm = 1.2 × 10-6m]
∴ Angular width of central maximum = 2θ = 2 × 30° = 60°
Example 4.
A Fraunhofer diffraction pattern is being formed by a light wave of frequency 5 × 1014Hz through a slit of width 10.2 m. Find the angular width of central maximum. (Velocity of light in vacuum = 3 × 108 m ᐧ s-1]
Solution:
The angular width on either side of the central maximum,
θ = \(\frac{\lambda}{a}\) = \(\frac{c}{\nu a}\) [∵ λ = \(\frac{c}{\nu}\)]
= \(\frac{3 \times 10^8}{5 \times 10^{14} \times 10^{-2}}\) [here, \(\nu\) = 5 × 1014 Hz, a = 10-2m, c = 3 × 108 m ᐧ s-1]
= 0.6 × 10-4 rad
∴ Angular width of central maximum = 2θ = 1.2 × 10-4 rad
Example 5.
A single narrow slit of width a is illuminated by monochromatic parallel ray of light of wavelength 700 nm. Find the value of a in each case following the given conditions-
(i) first minimum for 30° diffraction angle and
(ii) first secondary maximum for 30° diffraction angle. [CBSE ‘07]
Solution:
i) From the conditions of formation of n th minimum,
a sinθ = nλ
or, a sin30° = 1 × 700 × 10-9 [∵ θ = 30°, n = 1 and λ = 700 nm = 700 × 10-9 m]
∴ a = \(\frac{700 \times 10^{-9}}{\frac{1}{2}}\) = 14 × 10-7m
ii) From the condition of formation of n th secondary maximum,
a sinθ = (2n + 1)\(\frac{\lambda}{2}\)
or, a sin 30° = \(\frac{3}{2}\) × 700 × 10-9
[∵ θ = 30°, for 1st secondary maximum, n = 1 and λ = 700 nm = 700 × 10-9 m]
∴ a = 21 × 10-7 m
Example 6.
A star is observed through a telescope. The diameter of. the objective of the telescope is 203.2 cm. The wave length of the light, coming from the star to the telescope is 6600Å. Find the resolving power of the telescope.
Solution:
Resolving power of a telescope,
R = \(\frac{a}{1.22 \lambda}\) = \(\frac{203.2}{1.22 \times 6600 \times 10^{-8}}\) = 2.52 × 106
Here, the diameter of the objective of the telescope,
a = 203.2 cm
and wavelength of the incident light, λ = 6600Å = 6600 × 10-8cm
Example 7.
Find the Brewster angle for air to glass transmission. (R.I. of glass = 1.5) [WBCHSE Sample Question]
Solution:
If the refractive index of glass relative to air be μ and polarising angle be ip, then according to Brewster’s law we have,
μ = tan ip or, tan ip = 1.5 ∴ ip = tan-1 (1.5) = 56.3°
Example 8.
A single narrow silt of width a is illuminated by white light [Fig.]. For what value of a will the first minimum of a red light of wavelength 650 nm, lie at point P? For what wave-length of the incident light will the first secondary maximum lie at point P?
Solution:
From the condition of formation of first minimum at point P,
a sinθ = n λ
Here, n = 1, θ = 30° and A = 650 nm
a sin 30° = 1 × 650 or, a = 1300 nm
If first secondary maximum lies at point P, then
a sinθ = (2n + 1)\(\frac{\lambda^{\prime}}{2}\) or, a sinθ = \(\frac{3}{2}\)λ’
[here, λ’ = wavelength to be determined and n = 1]
or, λ’ = \(\frac{2}{3}\)asinθ = \(\frac{2}{3}\) × 1300 × \(\frac{1}{2}\) = 433.33
∴ The required wavelength = 433.33 nm
Example 9.
The refractive Index of glass is 1.55. What is its polarising angle? Determine the angle of refraction for the polarising angle.
Solution:
If the refractive index of glass relative to air be μ and polarising angle be ip, then according to Brewster’s law we have,
μ = tan ip or, tan ip = 1.55
∴ ip = tan-1(1.55) = 57.17°
Again for incidence at the polarislng angle,
ip + r = 90° [here r = angle of refraction]
or, r = 90° – ip = 90° – 57.17° = 32.83°
Example 10.
The critical angle of a transparent crystal is 30°. What is polarising angle of the crystal?
Solution:
If μ be the refractive index and be the critical angle of the crystal, then
μ = \(\frac{1}{\sin \theta_c}\) = \(\frac{1}{\sin 30^{\circ}}\) = 2
From Brewster’s law we have,
tan ip = µ [here ip = polarising angle]
or, tan ip = 2 or, ip = tan-1(2) = 63.43°
Example 11.
Determine the polarising angle of the light ray moving from water of refractive index 1.33 to glass of refractive index 1.5.
Solution:
From Brewster’s law we have,
Example 12.
When sun ray is incident at an angle 37° on the water surface, the reflected ray gets completely plane polarised. Find the angle of refraction and refractive index of water.
Solution:
According to question, angle of incidence
= (90° – 37°) = 53°
∴ Angle of polarisation, ip = 53°
Now, ip + r = 90° or, r = 90° – ip, or, r = 90° – 53° = 37°
According to Brewster’s law, refractive index of water,
μ = tan ip = tan 53° = 1.327