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 27

EASY

12th Meghalaya Board

IMPORTANT

Earn 100

What does it mean to be in the same phase of two waves?

Important Questions on Wave Optics

HARD

12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 28

The semi vertical angle of cone of the rays incident on the objective of microscope is

20 ° .

If the wavelength of incident light ray is

6600

\overset{°}{A}

. Calculate the smallest distance between two points which can be just resolved.

MEDIUM

12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 29

What is the validity of ray optics

?

HARD

12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 30

What are resolution and resolving power?

MEDIUM

12th Meghalaya Board

IMPORTANT

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

Write down main difference between Fresnel and Fraunhofer diffraction.

MEDIUM

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.

EASY

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_____.

EASY

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

MEDIUM

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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