Showing posts with label Interference. Show all posts
Showing posts with label Interference. Show all posts

Resolving Power of Optical Instruments

Resolving power of an optical instrument the is its ability to show to close lying objects as the separate entities in its image. This is different from magnifying power. It is not about increasing the size of the image. But it is about identifying the image of different bodies separately.

Consider a parallel beam of light falling on a convex lens. All the beam of the light rays are supposed to focus at a particular point. However due to the diffraction, instead of the beam pointing out at a particular point it is focused at a finite area. The image pattern consists of central bright region surrounded by concentric dark and bright regions as shown below.



Resolving power of a microscope

A microscope resolves the linear distance between two close objects. In the diagram shown the convex lens is having an aperture. Two different possible light rays at the two extremely ends forms two different images and identifying them separately is called resolving power. Resolving limit is the reciprocal of resolving power.




Resolving power of a telescope

A telescope gives resolution between two for away objects. We can explain that the resolving power of a telescope is directly proportional to aperture of the lens and inversely proportional to wavelength of the light used.




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Wave Optics Complete Lesson

Wave optics is a branch of physics which can explain the properties of the light like interference, diffraction and polarization. Here light is assumed to be travelling like a wave.

When two coherent sources of light are superimposed with each other, the resultant mate bright and dark spots formed and this phenomenon is called interference. The phenomena of the bending of the light at the obstacle are called diffraction. The phenomenon of restricting the light to a particular plane is called polarization. Here in this post all the details of all these topics are listed as shown below.





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Interference Fringe shift and Problems

Problem and solution

In the double slit experiment when a monochromatic source of wavelength lambda is used, the intensity of the light at a point on the screen for a path difference equal to the wavelength of the light is equal to K units. What is the intensity of the light at a point on the screen where the path difference is one third of the wavelength of the light?

When the path difference between the two sources is lambda, the resultant intensity is maximum. When the path difference is one third of the wavelength of the light we can calculate the corresponding phase difference. By substituting this value of the phase difference in the equation further resultant intensity we can get the value of resultant intensity as shown below.



Fringe visibility

Fringe visibility is the measure of comparison of the bright spots with respect to dark spots. We can write a small mathematical equation for it as shown below.



Fringes shift when a slab is placed in the path of the Ray

When a transparent sheet of thickness t is introduced in the path of the interfering wave, the interference pattern on the screen will shift to a different location.

We know that refractive index can be written like the ratio of apparent thickness the real thickness. Therefore because of the placing of the medium, there is a change in the path of the light as shown below. We can further simplify and identify the shift generated by the slab as shown below.



Problem and solution

In the double slit experiment how many maximums can be obtained on a screen when a wavelength of light 2000 Å is used and the distance between the slits is  7000 Å.

We can write the condition for the path difference as the integral multiples of wavelength of the light for the constructive interference. Basing on this condition we can solve the problem as shown below.



Problem and solution

In the double slit experiment when a monochromatic source of wavelength lambda is used, the intensity of the light at a point on the screen for a path difference equal to the wavelength of the light is equal to K units. What is the intensity of the light at a point on the screen where the path difference is one third of the wavelength of the light?

When the path difference between the two sources is lambda, the resultant intensity is maximum. When the path difference is one third of the wavelength of the light we can calculate the corresponding phase difference. By substituting this value of the phase difference in the equation further resultant intensity we can get the value of resultant intensity as shown below.



Problem and solution

In double slit experiment intensity at a point these one by fourth of the maximum intensity. What is the angular position of this point on the screen?

We can calculate the phase difference between the two points to have intensity one by fourth of the maximum intensity. It is found that the phase difference is 120°. By calculating the corresponding path difference and writing in the appropriate formula we can get the location of the point on the screen as shown below.



Problem and solution

In double slit experiment two slits are separated by 0.25 cm and the screen is it 120 cm from them. The slits are eliminated by light of wavelength of 600 nm. What is the distance between the first point on the screen from the central maximum where the intensity is a 75% of the maximum intensity?

As the intensity at a given point these three by fourth of the maximum intensity, we can calculate the phase difference between the two points as 60° as shown below. We can further calculate the path difference and by substituting the value of the path difference on the location of the point, we can calculate it as shown below.



Problem and solution

A double slit experiment is performed in a liquid. The 10th bright fringe in liquid lies where the six the dark fringe lies in vacuum. What is the refractive index of the liquid?

Basing on the mathematical equations that we have derived, we can identify the position of the bright spot as well as the dark spot. By substituting the given values in the problem we can get the answer as shown.

The attached paper is having one more problem and the detailed solution is also given. For any further clarifications it can comment at the end of the post.





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Fringe Width and Angular Fringe Width in Interference

Fringe width is the distance between two successive bright fringes or two successive dark fringes. In the interference pattern, the fringe width is constant for all the fringes. It means all the bright fringes as well as the dark fringes are equally spaced. We can derive the equation for the fringe width as shown below.

Fringe width is independent of order of fringe.

Fringe width is directly proportional to wavelength of the light used.

Hence fringe width for the red colored source is going to be greater than the fringe width of violet colored source.

If the interference experiment is conducted in the denser medium, the corresponding fringe width is reduced by its refractive index times. It means when the experiment is conducted in the denser medium we can observe that the fringes appear close to each other.

We can also define the angular fringe width as shown below.



In the interference pattern the bright spots and the dark spots are equally spaced. The path difference as well as the phase difference between any two successive bright spots or any to the successive dark spots is always constant. The variation of intensity of the output wave is regular and systematic as shown below.



Problem and solution

Two coherent sources are 0.18 mm apart and the fringes are observed on the screen 80 cm away. It is found that the fourth bright fringes 10.8 mm away from the central bright fringe. What is the wavelength of the light?

We know the condition for the constructive interference and the location of the bright spot on the screen. Using that formula we can find out the wavelength of the light as shown below.



Problem and solution

In the double slit experiment the slits are separated by 0.1 mm and they are at 50 cm from the screen. The wavelength of the light used these 5000 Å. Find the distance between seventh maximum and 11 minimum on the screen.

The formation of bright spots in the dark spots on the screen due to interference pattern is regular and systematic. We can apply the formula that we have derived for a location of the bright spot and the dark spot as shown below and solve the problem.




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Interference of Light an Overview

The variation of intensity in the region of superimposition of two or more than two waves of same frequency with a constant phase difference is called interference.If two waves are met in such a way that the resultant intensity is maximum, then it is called constructive interference. This is possible when the two waves are met in the same phase.

If the two waves are met in such a way that the resultant intensity is minimum, then it is called destructive interference. This is possible when the two waves are met in the opposite phase.

For producing interference pattern of light, the two sources shall be coherent. Two sources are said to be coherent when there have a zero phase difference or a constant phase difference. Two different sources of light will never be coherent. We can get two coherent sources only when both the sources are drawn from a single source. A source and its image also can behave like coherent sources. The image can be a real image or it can be a virtual image.

An experiment is conducted to produce interference pattern and it is called Young’s double slit experiment. The light from a source is allowed to pass through two small slits. These two slits acts like coherent sources. They also behave like secondary sources with a constant phase difference. Hence they are qualified to produce interference pattern.

For interference pattern to appear on the screen, the distance between the slits and the screen shall be much larger than the distance between the slits.



We can draw a diagram to represent constructive interference and the destructive interference. When the two waves from the different sources are met in the same phase, the resultant intensity is maximum and they produce a bright spot. This is called constructive interference. When the two waves are met in the opposite phase, the resultant intensity is minimum and they produce a dark spot. This is called destructive interference. The central spot is a bright spot because by the time the two waves reach the point, there have zero phase difference.



To know the resultant intensity and the resultant amplitude at a particular point on the screen, we can draw the light from different sources at a point on the screen as shown. The two lights while reaching the point will experience a small path difference and hence there will be a constant phase difference also between them. They are represented in the diagram as shown below.



When the two waves are superimposed the resultant amplitude is different from the individual amplitudes and it also depends on the phase difference between the two waves. We can derive the equation for resultant amplitude and resultant intensity as shown below.




The formation of bright and dark fringes in interference pattern depends on how the two waves are met at a particular point. It can be proved mathematically that when the two waves are having a constant path difference that is equal to wavelength of the light, the resultant intensity is going to be maximum and that is called constructive interference. We can derive the equation for the constructive interference and the location of the bright spot on the screen as shown below.




The formation of the dark spot on the screen is possible when the two waves are met with a constant path difference of half of the wavelength. We can derive the condition as shown below.



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Representation of Light as Wave

Representation of the light as a wave

We can consider the light travelling like a wave. The particles of the medium through which the light is passing experience a displacement and it can be represented as shown below. Even before the oscillation starts the particles can have some displacement and it is called as initial phase. The position of the particle with respect to the mean position is called phase. The wave moving along the positive direction and the wave moving along the negative direction are shown with a different signs as shown below. We have also derived a small relation between phase difference and path difference.



We also know that velocity of the wave is the product of frequency of the wave with its wavelength. We shall understand that the velocity of the particle is different from the velocity of the wave.

The intensity of the light at any point is directly proportional to Squire of the amplitude.




Principle of superposition

When two waves are superimposed one over each other, the resultant of displacement is going to be different from individual displacements.

Treating the displacement as a vector, we can calculate the resultant amplitude as shown below. The derivation is made basing on the parallelogram law of vectors. It is clear from the derivation that the maximum possible amplitude of the two waves is equal to sum of the individual waves amplitudes. The minimum possible amplitude of the resultant wave is equal to the difference between the amplitudes of the two waves.

We can calculate the ratio of maximum amplitude to minimum amplitude as shown below. As we know that the intensity is directly proportional to Squire of amplitude we can also calculate the ratio of maximum intensity to minimum intensity.




Doppler effect of light

The apparent change in the frequency due to the relative motion is called Doppler effect. The change in the apparent frequency is not dependent of change in the velocity of the observer. It is simply because when compared with the velocity of the light, the velocity of the observer is significantly small. Therefore the impact of motion of the observer is less on the apparent change in the frequency of light.

We can explain the concept of blue shift and the red shift basing on Doppler Effect of light. When an astronomical body is approaching the earth, its apparent frequency increases. We know that the wavelength is reciprocal of frequency. As the frequency of the approaching body is increasing, its wavelength decreases. Among all the visible colors, violet is having the least possible wavelength but it is not a primary color. As the closest color with the dominating wavelength is blue, the body approaching the earth appears in blue color.

If an astronomical body is going away from the earth, its wavelength increases and it appears like red in color. This is called the red shift of the star.




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