Show using a proper diagram how unpolarized light can be linearly polarised by reflection from a transparent glass surface.

When an unpolarized ray of light strikes upon a transparent glass surface, the reflected light gets polarised. But when the reflected and refracted light are perpendicular to each other, the ray of light gets completely plane polarised. Thus, if i p and r be the angle of incidence and refraction, when reflected and refracted light is perpendicular to each other respectively, then: i p + 90 ° + r = 180 ° Using Snell's law: μ = sin ( i p ) sin ( r ) Using the result from above, μ = sin ( i p ) sin ( 90 ° - i p ) μ = sin ( i p ) cos ( i p ) = tan ( i p ) This is the equation for Brewster's law. The angle of incidence i p is called Brewster's angle. Thus, when an unpolarized light is an incident on a transparent glass surface at Brewster angle, the reflected light is plane-polarised light. The diagram for the same is given below:

Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for maximum and zero intensity.

Let the two waves S 1 and S 2 from the two coherent sources interfere at a point P. Let their amplitudes be a and the initial phase difference between them be ϕ . Then the displacements produced by S 1 and S 2 will be given as: y 1 = a cos ω t ....(i) y 2 = a cos ωt + ϕ ....(ii) The intensity produced by each wave, I 0 ∝ a 2 ....(iii) The resultant displacement at point P will be: y = y 1 + y 2 y = a cos ω t + cos ω t + ϕ y = 2 a cos ϕ 2 cos ω t + ϕ 2 ......(iv) From (iii), it is clear that the amplitude of the resultant displacement is 2 a cos ϕ 2 and therefore the intensity at that point will be: I ∝ 4 a 2 cos 2 ϕ 2 On using (iii), we get: I = 4 I 0 cos 2 ϕ 2 .....(v) Where I 0 is the intensity of each of the individual sources. Using (v) , For ϕ = 0 , ± 2 π , ± 4 π . . . . . , the intensity will be maximum and given by: I = 4 I 0 For ϕ = ± π , ± 3 π , ± 5 π . . . . , the intensity will be zero: I = 0

E M B I B E

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 31

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12th Meghalaya Board

IMPORTANT

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Write down main difference between Fresnel and Fraunhofer diffraction.

Important Questions on Wave Optics

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12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 32

Describe with neat diagram, the Fraunhofer diffraction pattern due to single slit.

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12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 33

Two identical light waves having phase difference

ϕ

propagate in the same direction. When they superpose, the intensity of the resultant wave is proportional to_____.

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12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 34

In the interference pattern by two identical slits, intensity of central maxima is I. If one of the two slits is closed, then intensity of light at the same spot will be

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12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 35

In Young's double-slit experiment. If the distance between the two slits is equal to the wavelength of used light. Then the maximum number of bright fringes obtained on the screen will be

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12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 36

In two separate setups of Young's double slit experiment, fringes of equal width are observed when lights of wavelengths in the ratio 1 : 2 are used. If the ratio of slit separation in two cases is 2 : 1, then ratio of distance between the plane of slits and the screen in the two setups is

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12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 37

Light of wavelength