Showing posts with label Wave Optics. Show all posts
Showing posts with label Wave Optics. 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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Diffraction in Wave Optics Overview

The phenomenon of bending of the light and the obstacle is called diffraction. It is exhibited by not only the light waves but also by sound waves, Matter waves and water waves. Because of the diffraction effect of sound waves, we are able to hear the sounds made by the other people that are there in a different room. It is just because sound bends at the Windows.

The encroachment of the light into the darker regions is expressed as diffraction. If an obstacle is placed in the path of the light, the entire corresponding region is supposed to become dark. But because of the encroachment of the light, even the supposed to be darker region consists of bright and dark bands. The formation of the bright and dark bands is due to mixing of the waves that are coming from the same source that are split up into multiple sources.



The diffraction pattern consists of alternate bright and dark bands. But the intensity of All bright bands as not equal. The bright bandit the Centre has maximum brightness and it is called principal maximum. The dark bands surrounding the bright band are called as minima.

The other bright bands have decreasing intensity away from the principal axis and they are called secondary maxima.

As the bands progresses, bright bands becomes less bright and the dark bands becomes less dark. It means after a few bands will become difficult to identify that who is a bright band and who is the dark.



We can derive a small mathematical equation for minimum and maximum in a diffraction patron as shown below. Here we have only one source and when it is restricted by the obstacle it behaves like multiple sources. The resultant intensity is being observed on your screen. We can write equation for the bright and dark spots as shown below.




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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.





Polarisation Concepts and Problems

The phenomena of restricting the vibrations of the light to a particular plane is called polarization. It is the property of the transverse wave. A transverse waves can only have vibrations in both the planes. Hence restricting the vibrations of one plane is possible with the transverse wave.

Sound waves in air cannot be polarized because they’re longitudinal in nature.

To explain the concept of polarization we shall depend on an electromagnetic wave theory of light. According to this theory, the varying electric fields particles generate a varying magnetic field around it and the direction of wave motion is perpendicular to both of them.

The magnetic field vibrations are no way responsible for any of the optical properties of light. Electric fields vector is responsible for all optical properties of light. This electric field vector can be resolved into components. The horizontal and vertical components exist in the perpendicular planes. The light that is having both these components is called unpolarized light or natural light. To polarize the light, we shall restrict either horizontal vibrations are vertical vibrations. Horizontal vibrations are represented with arrows and the vertical vibrations are represented with darts.

A unolarised light consists of both the horizontal and vertical vibrations. When the light is polarized, it will have either only the horizontal vibrations or only the vertical vibrations.

The intensity and the amplitude of the polarized light is less than that of the original light. To be more specific intensity of polarized light is half of the original intensity.
Polarization can be done by certain type of crystals like quartz crystal. This crystal is called as polarizer. To check whether the light is actually polarized or not, we can use the same crystal and then it is called as analyzer. Depending on the position of analyzer, the outcome of the light and its intensity can be identified as shown in the diagram.



Problem and solution

Unpolarised light of intensity that the 32 W per meter Squire passes through three Polaroid’s such that the transmission axis of the last polariser is crossed with the first one. If the intensity of the emerging light is 3 W per meter Squire what is the angle between the axes of the first two?

It is given in the problem that the sum of the two angles is equal to 90°. We know the equation further resultant intensity and by writing in this equation and simplifying we can get the answer as shown below.



Problem and solution

A beam of ordinary light is incident and system of four Polaroid’s which are arranged in succession such that each Polaroid is turned through 30° respect to the preceding one. What is the percentage of the incident intensity that emerges out from the system?

We know that when the light is passed through Polaroids its intensity decreases. If it is passing through only one polariser with the parallel axis only half of the intensity is going to remain. When there are good number of the polarisers, to find out the resultant intensity we can use a formula that is shown below and get the intensity.



Problems and solutions

In the following page attached, three problems are solved with a detailed solution. A comment can be made at the end of the post for any kind of further clarifications?



Polarization can be achieved through the process like reflection, refraction and double refraction. When the light strikes the boundary that is separating the two media, for a particular angle of incidence the reflected light is completely polarized. This particular angle of incidence is called the angle of polarization. In this case the refracted light consists of a small number of vertical vibrations and hence the refracted light is not completely polarized.

We can nature the polarization even by refraction. Here a group of plates are arranged one after the other so that each plate to its own polarization up to some extent. As many number of the required plates are placed such that the refracted light is having only horizontal vibrations. This combination is called pile of plates.

We can get the polarization by double refraction also with the help of so special crystals like a calcite crystal. This crystal has a specific property of reflecting the light twice and this property is called the double refraction. It is experimentally observed that both this refracted light’s are completely polarised.

One of the refracted lights follows all the regular rules of light and it is called ordinary light Ray. With the ordinary light Ray refractive index of the medium and the velocity of the light is constant.

The second refracted light Ray doesn’t follow the regular rules and it is called the extraordinary light Ray.

We can identify the ordinary and extraordinary light Ray quite easily. When the calcite crystal is placed in on a point object it makes to point images. With the rotation of the crystal one image reminds in the state of rest and the image is as ordinary. With the rotation of the crystal other image starts rotating and it is called the extraordinary.




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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 Ã… is used and the distance between the slits is  7000 Ã….

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 Ã…. 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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