Test: Wave Optics- 2 (March 5) - NEET MCQ

white light is made up of seven colors VIBGYOR. All the seven colors will show interference, we can treat each color separately as a monochromatic light and study their interference separately. Each light will have their bright and dark fringes separately but at y=0 i.e for a center, each lights bright fringe will coincide. Two or more lights may have their bright and dark Fringe together or bright for one and dark for others may coincide depending upon the situation.

When Young's double-slit set up for interference is shifted from air to within water then the fringe width decreases because the refractive index of water is more than that of the air.
Originally the fringe width is given by:
β1​=λD/2d​
The new fringe width within water will be given by 
β2​= λD​/2nd
So, β2​= β1/n​​
Here, n is the refractive index of medium.
 

Dark fringes will be produced when there are destructive interference. The condition for that is the two waves should have a phase difference of an odd integral multiple of π.

Fringe width β=λD/d​
On d′=d/2​, D′=2D
New fringe width β′= λD′​/d′=4β

We observe interference pattern when the size of the slits are comparable to the wavelength of light. But here, the size of the bulb is bigger than wavelength of light.

 Suppose the angle between the transmission axes of the analyser and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyser. If E₀ is the amplitude of the electric vector transmitted by the polarizer, then intensity I₀ of the light incident on the analyser is
I ∞ E₀²
 
The electric field vector E₀ can be resolved into two rectangular components i.e E₀ cosθ and E₀ sinθ. The analyzer will transmit only the component ( i.e E₀ cosθ ) which is parallel to its transmission axis. However, the component E₀sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is,
 
I ∞ ( E₀ x cosθ )²
 
I / I₀ = ( E₀ x cosθ )² / E₀² = cos2θ
 
I = I₀ x cos²θ
when light passes from polarizer it's intensity becomes half and when passed through analyser it becomes,
I = I₀ x cos²θ/2
 

Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized.
So the angle of incidence would be 60°

An analyser or analyzer is a person or device that analyses given data. It examines in detail the structure of the given data and tries to find patterns and relationships between parts of the data. An analyser can be a piece of hardware or a computer program running on a computer.

The polarization angle (also called Brewster's angle)is defined as the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.
Brewster's Law states that the tangent of the angle of polarization is numerically equal to the refractive index of the medium.
μ = tan i ...................... (1)
From Snell's law:
μ = sinr sini​ ....................(2)
Comparing (1) and (2)
cos i = sin r = cos (2π​ - r)
i.e. i = (2π​ - r)
So, i + r = 2π​
So, as seen from the diagram, if i + r = 90^o Then the angle between reflected ray and refracted ray will be 180^o - 90^o = 90^o

According to Malus’s law Iαcos²θ, when θ=0^o or 180^o, I=I_ocos²0^o = I_o
i.e the intensity of the light transmitted by the analyser is maximum when the transmission axes of analyser and polarizer are parallel.