What is a polarisation of light? Explain pillarization of light by reflection with the suitable diagram and hence derive Brewster's law.

Polarisation of light is a property shown by transverse waves. The light waves which travel only in a single plane are known as polarised light waves. The process of transforming unpolarized light waves to polarised light waves is called the polarisation of light. It is a property applying to transverse waves that specifies the geometrical orientation of the oscillations Polarisation By Reflection: If light incidents on a refractive surface so that the refracted light and the reflected light are at 90 ° angle, then the reflected light will be linearly polarised, and the plane of polarisations will be parallel to the plane of the interface. As per the Brewster’s Law, the maximum polarisation takes place when the angle is 90 degrees in between the reflected ray and refracted ray. The polarizations angle is called Brewster’s angle. In this case μ = tan I p where I p = Brewster 's angle Refer to the figure , when light is incident at polarising angle p on a refracting medium of refractive index μ , Let r be the angle of refraction, then according to the Snell's law. μ = sin I p sin r and from the brewster's Law sin I p sin r = tan p or sin p sin r = sin p cos p or sin r = cos p or sin r = sin ( 90 - p ) or r = 90 - p or r + p = 90 . Hence, when a ray of light incident at the polarising angle, the reflected ray is perpendicular to the refracted ray. This is Brewster's law.

In Young's double-slit experiment, two slits o . 125 mm apart are illuminated by the light of wavelength 4500 ° A . The screen is 1 m away from the plane of the slits. Find the separation between second bright fringes on both sides of central maxima.

Given the separation between the slit d = 0 . 125 mm = 0 . 125 × 10 - 3 m The wavelength of the light λ = 4500 ° A = 4500 × 10 - 10 m . The distance of the screen D = 1 m . No. of fringe n = 2 If x is the distance of the second fringe on each side, then the separation between them is, 2 x = 2 nλ D d . Substituting the values, we get 2 x = 2 × 2 × 4 . 5 × 10 - 7 × 1 0 . 125 × 10 - 3 = 144 × 10 - 6 m . Hence, the separation of second bright fringes is 14 . 4 mm .

State two conditions for two light sources to be coherent.Derive an expression for the fringe width in Young's double slit experiment for interference of light waves with suitable diagram.

Following are the conditions for the two lights to be coherent: (i) They should be of the same frequency. (ii) They should be preferably of the same amplitude. Consider two coherent sources of light S 1 and S 2 are placed at a distance d apart and a screen is placed at a distance D from the two sources. Let P be a point on the screen at a distance y from the centre of the screen. If x is the path difference between the light waves then, x = S 2 P - S 1 P In right-angled ∆ S 2 B P , we have S 2 P 2 = S 2 B 2 + B P 2 = D 2 + y + d 2 2 Also in right-angled ∆ S 1 A P , we have S 1 P 2 = S 1 A 2 + A P 2 = D 2 + y - d 2 2 Thus, S 2 P 2 - S 1 P 2 = D 2 + y + d 2 2 - D 2 + y - d 2 2 or ( S 2 P + S 1 P ) ( S 2 P - S 1 P ) = y + d 2 2 - y - d 2 2 Since, S 2 P - S 1 P = x (the path difference) Thus, ( S 2 P + S 1 P ) · x = 4 y · d 2 = 2 y d or x = 2 y d S 2 P + S 1 P In practice, S 2 P ≈ S 1 P ≈ D Hence, x = 2 y d 2 D or x = y d D . This is the expression for path difference. The point P will be a centre of bright fringe, if y d D = n λ or y = n λ D d , where, n = 0 , 1 , 2 , 3 , . . . . . . . Similarly, the point P will be a centre of dark fringe, if y d D = ( 2 n + 1 ) λ 2 or y = ( 2 n + 1 ) λ D 2 d , where, n = 0 , 1 , 2 , 3 , . . . . . . . The spacing between any two consecutive bright fringes or dark fringes is called fringe width. It is represented by β . Thus, β = y n - y n - 1 = n λ D d - ( n - 1 ) λ D d = λ D d (for bright fringe) and β ' = y n ' - y n - 1 ' = ( 2 n + 1 ) λ D 2 d - 2 n - 1 + 1 · λ D 2 d = λ D d (for dark fringe) Thus, it is clear that, β = β '

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 10 - Wave Optics>Punjab Board-2018>Q 2

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12th Punjab Board

IMPORTANT

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What is the essential condition for diffraction of light to occur ?

Important Questions on Wave Optics

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12th Punjab Board

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 10 - Wave Optics>Punjab Board-2019>Q 4

A parallel beam of light of wavelength $500 nm$ falls on a narrow slit and the resulting diffraction pattern is observed on a screen $1 m$ away. It is observed that the first minimum is at a distance of $2.5 mm$ from the centre of the screen. Find the width of the slit.