## What is Physical Optics?

In physics, physical optics, or wave optics, is a branch of optics that studies interference, landscaping, polarization, and other phenomena for which it is not correct to approach a ray of geometric theory. This use does not include effects such as quantum noise in optical communication, which is studied in the sub-branch of the theory of compatibility.

### Physical Optics

Physical optics includes effects such as the propagation of light in a medium such as the optical waveband, as well as vision phenomena such as visible light, colors, and our standard model of vision. Applications In addition to vision, physical optics and its analogues are important for radar, optics and lasers, acoustics, medicine, and light localization. The theory of physical optics requires the explicit solution of diffraction, dispersion, and reflection equations in a given medium, and their integration over arbitrary sources and media. There are many practical problems in optics that are understood in terms of a symmetry group, such as lensing. Density functional theory (DFT) is used in theoretical particle physics.

### Why is physical optics important?

Physical optics is relevant to medicine, since only a small part of wave propagation is predictable in optics, especially with the use of lasers. When people aim lasers, the direction of the scattered beam is always unknown and non-linear effects can be important. Physical optics is an integral part of laser light. Although the accuracy of 3D measurements with conventional instruments such as high accuracy parabolic microphones (AOI) is good, the effect of particle size and momentum is nonlinear, so the errors in dimensions and time vary a lot from spot to spot and can be from a few millimeters to thousands of kilometers. Physical optics analysis is a way to overcome this error to give a better understanding of the phenomenon.

### Applications of Physical Optics

Electromagnetic waves (light, sound, etc.) are commonly described by propagation of frequencies and phases, in particular interference between these. Optical signals and modulations of these signals can be used to define digital information and control systems, making it the next target of practical applications of physical optics in information processing. Physical optics can be used for image projection or controlling light. Applications of Physical Optics: Switching and Switching Control The control of switching and switching control is called physical optics and the switching process is modeled by a combination of light propagation and Mach–Zehnder interferometry.

#### Conclusion

1) Real physical optics is nothing like vision 2) “Bezels and Siblings” in digital cameras 3) Touch screens 4) Quantum noise in our fiber optic cable system 5) Inability of humans to track each other with any accuracy 6) Our natural biological body movement has many distortions and inaccuracies 7) Our brains are hard-wired for stationary vision 8) Multi-touch screens 9) Optical cable system 10) First-person shooter games 11) Multiple distortion (light vs. dark) 12) Three-dimensional imaging system 13) Controlling the electron beam in lenses with more sensitivity than the electron beam 14) Reading/writing bytes with electromagnetic fields 15) Different camera viewing and optical composition systems (photographic vs.

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### A soap bubble looks black when it bursts, why?

Soap bubble breaks when its thickness becomes almost zero, such that the path difference between the upper surface and lower surface becomes zero. If the light beam falls on the bubble then its upper surface will reflect the beam as it is denser as compared to the lower surface. So, for the first ray, there will be a path difference of λ/2.

Now as the second ray will be reflected from a rare medium i.e., from the lower surface so it, there will be no path difference at all.

Therefore, the total path difference between the 1st and 2nd rays will be λ/2 which results in destructive interference of light wave. That is the reason that the soap bubble looks black when it is about to break.

### What is the difference between interference and diffraction?

Interference is the superposition of two waves, while diffraction occurs when waves bend around the slit edges when they travel.

Diffraction fringes can never be of same width, whereas interference fringes may or may not be of same width.

Intensity of interference fringes is high as compared to diffraction fringes.

In interference, the areas where the intensity is low will be perfectly dark but bright areas will have uniform intensity throughout. However, in diffraction points of low intensity are dark but not with perfection and neither the bright areas have uniform intensity as they do in case of interference.

### In a Michelson interferometer a second glass plate is also used, why?

In Michelson interferometer, a second plate is used to make a constructive interference so that one might be able to get bright fringes with the help of the telescope.

Constructive interference occurs due to the zero-path difference between two beams as the first beam passes through the first plate and second beam pass through the second plate. So, in that way, the second plate produces the same retardation as the first plate does.

**How you can explain Brewster’s law of polarization?**

Brewster’s law gives us the relation between the ratio of refractive indexes of two mediums and angle of polarization. **Mathematical Expression: **

**Derivation:**

Consider an unpolarized beam of light is falling on glass or water. By doing so, light will interact with two mediums 1 and 2, say air and water, respectively. At point of incident, one part of light will be reflected and get polarized completely, while the other part will be refracted from the surface and will be polarized partially.

If ‘i_{P}’ be the angle of polarization and ‘r’ be the angle of refraction. The angle at which the polarization is complete, the reflected ray and refracted ray will be at right angle to each other in the transmitted medium. The vibrations in reflected ray are parallel to the surface of the water, then according to Snell’s law we have

n_{1} sin i_{P} = n_{2} sin r ……..………………………….. (1)

Where n_{1} and n_{2} are absolute refractive indexes of medium 1 and 2, respectively. From figure, we can write

i_{P} + 90^{0} + r = 180^{0}

⇒ r = 180^{0} – 90^{0} – i_{P} = 90^{0} – i_{P}

Putting this value in equation (1), we get

n_{1} sin i_{P} = n_{2} sin (90^{0} – i_{P}) = n_{2} cos i_{P}

**Q.5) What is meant by the path difference with reference to the interference of two wave motion?**

**Answer:**** Interference:**

The effect produced by the superposition of waves from two coherent sources passing through the same region”**Path Difference:**

If two waves are passing through a region with same velocity, frequency and wavelength then they will traverse the same length. Therefore we say that they have same phase difference. But if one of the wave will travel a longer distance than the other but have the same speed, frequency and wavelength then there will be a difference in their path length.

This difference is known as the path difference. It is the path difference of two waves which decides whether the interference will be constructive or destructive.

For constructive interference,

Path Difference = P.D = mλ

Where, m = 0, 1, 2, 3, ……….. an integer

For destructive interference,

Path Difference = P.D = (m + ½)λ

Where, m = 0, 1, 2, 3, 4, ……

**Q.6) Why it is not possible to see the interference where the light beams from the headlamps of a car overlap?**

**Answer:**

The waves coming from the two lamps of car have a phase difference between them. This phase difference is not constant.

As for a wave to be coherent, the phase difference should be constant. Therefore, the light beam coming from the lamps of the car are not coherent and also not monochromatic. These are the reasons that it is not possible to see the interference where the light beams from the headlamps of a car overlap.

**Q.7) A telephone pole casts a clear shadow in the light from a distant head lamp of a car, but no such effect is noticed for the sound from the car horn. Why?**

**Answer:**

Both, the sound and light travel in form of waves. But the sound waves have a larger wavelength as compared to light waves. Therefore due to the larger wavelength of the sound waves, they bend around the edges of the pole and are heard.

But on contrary, the light waves do not bend across the pole as they do not have a comparable large wavelength. So that is the reason that we cannot see the light around the corners of the pole.

**Q.8) Why it is not possible to obtained the diffraction of X-rays by Young’s double slits experiment?**

**Answer:**

In Young’s double slit experiment, two slits are placed in the way of light waves to produce the interference.

X-Ray are electromagnetic waves who have very short wavelength as compared to the visible light. Therefore in order to get diffraction from them we cannot use ordinary diffraction objects and equipment. Thus by using ordinary slits, we cannot produce diffraction pattern of x-rays.

**Q.9) Can we apply Huygen’s principle to radar waves?**

**Answer:**

Yes, Huygens principle can be applied to radar waves. In radar waves, we deal with the propagation of waves and Huygens principle give us information about the propagation of the wavefronts from one point to another point in space.

**Q.10) How would you justify that light waves are transverse?**

**Answer:**

As only in transverse waves, polarization can take place. Therefore we can say that polarization explains that light waves are transverse in nature.

As there is no polarization occur in case of longitudinal waves so one can claim that light waves are transverse in nature as they exhibit polarization effect.

## Physics Class 11 2021 comprehension question Chapter 9

**Q.1) What is meant by the dual nature of light? Discuss the history of the nature of light in detail.**

** Answer:**

**Dual Nature of Light:**

Light has a dual nature, as it behaves both like waves and as well as particle-like. In the case of reflection, rarefaction, diffraction, interference, and polarization, light behaves like a wave.

But in the case of photoelectric effect and Compton shift, it has a particle-like nature.

Therefore, we say that light has a dual nature.

**History about the Nature of Light:**

Due to the dual nature of light, it was hard for scientists at the start to explain its nature i.e. whether it has a particle-like nature or if it is wave-like. Therefore, they make assumptions about the nature of light and develop different theories about the nature of light. Some of these theories are as follows:

**Maxwell Theory of Light:**

Maxwell proposed this very important theory in 1873. He showed that light is a form of electromagnetic waves which have a very high frequency i.e. it consists of both, electric vector and magnetic vector at the same time. These vectors oscillate perpendicular to the direction of the propagation of the wave. He predicts in his theory that the velocity of these waves is about 3 × 10^{8} ms^{-1}. He tells us that these waves do not need any type of medium to propagate.

**Huygens Principle:**

Huygens explains in his theory that light travels from one place to another in the form of waves. He explains the process of reflection and rarefaction in his theory. But the theory was failed as it gives no information about the process of polarization, diffraction, and interference of light.

**Young’s Explanation:**

In 1801, Thomas Young experimented. In his experiment, he explains the diffraction phenomena of light.

**Q.2) Explain the diffraction of X-rays by the crystal and derive an expression for Bragg’s law to find the wavelength of light used?**

** Answer:**

**Diffraction of X-Rays by Crystal:**

The atomic structure in crystalline objects is very important by which we can define different useful properties of the crystal. The atomic structure of crystalline objects is studied with the help of diffraction. As the atoms in the crystal layer are 1 nm apart i.e. much closer to each other. Therefore, to diffract a wave between two atoms must be of very short wavelength. X-rays are best for this purpose as they have a wavelength of order 10^{-10} m. X-rays interface fringes cannot be observed in Young’s double-slit experiment because the fringes are so close to each other in that case that we cannot observe. As the distance d is also very small in case of crystal so the crystal interface fringes are also apart from each other.

**Bragg’s Law: **

Consider two parallel incident rays I and II falls on a crystal’s surface and they reflect. Ray I incident on the first layer and ray II on the second layer. Let separation between two layers is ‘d’ while ‘θ’ is the glancing angle which is complementary as shown in the figure.

After interacting the crystal, the rays will refract and will reinforce each other if the path difference between the two incident rays is equal to λ or multiple of λ. Ray II covers longer distance as compare to Ray I, so the path difference is gives as

BC + CB’ = mλ …………………….. (1)

From figure, we can see that

BC = CB’=dsinθ

So putting this in equation (1), we get

dsinθ + dsinθ = mλ

⇒ 2dsinθ = mλ

Where m = 1,2,3,4…….

Which is known as Bragg’s Law. By the help of Bragg’s Law we can find the interplanar distance between the parallel planes of crystals when the wavelength of diffracted wave is known. It is also useful in determining the structure of Biological important molecules like Haemoglobin and DNA.

**Q.3) Describe the experimental arrangement for the production of interference fringes by Young’s double slits method, and get an expression for the fringes space.**

**Answer :Experimental arrangement: **

Young’s double slit experiment is a very famous experiment which is proof for the wave nature of light. In this experiment, Thomas Young split a monochromatic light and get its interference pattern. The experimental arrangement is shown in fig.

Figure shows that the monochromatic light from the source passes through a slit ‘C’ and falls upon slit A and B. slit A and B are at the same distance from slit C and they both act as two coherent sources of light. The screen is at distance D from slit A and B.

According to Huygen’s principle, the light passes through slit C so as the slit A and B are the same distance from C so the same wavefront arrives at A and B. so slit A and B also acts as two coherent sources and this method is known as “division of wavefront”. The slits are very small therefore diffraction occurs at both slit A and B and for the waves emerging from slit A and slit B interference of waves occurs and in the result of constructive and destructive interference, a pattern of dark and bright fringes appear on the screen which is at a distance D from the slits. The constructive interference results in bright fringes and destructive interference results in dark fringes.

**Fringe spacing:**

Fringe spacing is the distance between two consecutive bright or dark fringes.

So in case of bright fringes, the fringe space between 1st order and 2nd order bright fringes is

As we know for m^{th} bright fringes we have

As we know for mth dark fringe we have,

**Q.4) State and explain Huygen’s principle. What is the difference between spherical and plane wavefronts?**

**Huygen’s Principle:**

This principle states that,

“Every point of a wave front may be considered as a source of the secondary spherical wavelet, which spread out in forwarding direction with a speed equal to the speed of propagation of the wave. And the new position of the wavefront after time ‘T + Δt’ can be found by drawing a plane tangential to all the secondary wavelet”

If we know that shape and location of a wave front then by using Huygen’s Principle at any instant of time ‘t’ we can determine the shape and location of the new wavefront at the later time t+Δt.

**Diagram:**

**Explanation:**

Consider a source ‘S’ produces a wavefront ‘A’ as shown in the figure. Here AB represents the position of a spherical wave front a time ‘t’. We have to determine the new position and shape of wave front AB after time ‘t + Δt’. In order to do that following Huygen’s principle, we take several dots say 1, 2, 3, ….. ,10 on the wave front. Then spherical waves emitted from these points are shown in the figure by hemispheres of radius ‘cΔt’. Where ‘c’ is the speed of light here.

Now according to second part of the Huygen’s principle draw a surface A′B′ such that this surface is tangent to the secondary wavelet. Therefore, A′B′ is the new position of the wavefront after time Δt as shown I part (a) of the figure.

**WaveFront:**

“The locus of all the points in a medium which have the same phase is known as a wavefront”

There are two types of wavefronts:

**1. Spherical WaveFront**:

A wavefront which has concentric spheres such that the centre of all these spheres is their source then such wavefront is known as a spherical wavefront.

**2. Plane Wave Fronts:**

For spherical wavefronts, at a very large distance, they appear as planes. These straight parts of the wave front are known as plane wavefronts. These planes are usually parallel to each other.

**Q.5) Explain the interference effect produced by thin film.**

**Answer :Interference in Thin Films:**

Consider a thin film of a refracting medium with a thin wedge-shaped structure and refractive index. Let a monochromatic beam of light whose wavelength is ‘λ’ is incident on this thin film. On interacting with the film, it will split into two parts. One which is reflected from the upper surface say ‘part a’ and others which reflect from the lower surface say ‘part-b’. For a very thin film, the separation between the upper and lower layer is very small. Then both reflected beams will have phase coherence and they will superpose each other.

Path difference of both reflected beam depends on thickness and nature of the film and on the angle of incident. Interaction of a light beam with a thin films can be of two type as follows:

(a) If wave travels from a medium of lower refractive index to a higher refractive index medium then after reflection, it will has a phase change of 180^{o}.

(b) If wave travel from higher refractive index medium to a lower one then there be no phase change between incident and reflected beam.

**Q.6) What is the principle of interference of light? Discuss the necessary condition for interference of light**

**Answer :Interference of Light:**

“It is the effect produced by the superposition of waves from two coherent sources passing through the same region”

When two or more waves propagate in same direction through a certain region of space such that waves have same frequency, amplitude and a constant phase relationship then they will reinforce each other at some point and will cancel the effect of each other at some other point. Such phenomena of superposition of waves is called interference of waves.

**Conditions to observe Interference of Light:**

Following are the conditions which are required for the light waves to produce the effect of interference:

1. The light waves must come from two coherent sources.

2. The amplitude of the waves must be equal or nearly equal.

3. The light waves should be perfectly monochromatic.

4. The path difference of the waves from the two sources must be small.

5. The principle of linear superposition should be applicable.

**Q.7) What is diffraction grating? How can the wavelength of a beam of light be measured with it?**

**Answer :****Diffraction grating:**

A device which separates a single beam of light into several different parallel beams or lines on the bases of different wavelength is known as a Diffraction grating. The separation between these lines is known as Diffraction element.

Measurement of Wavelength:

Let a beam of light pass through a diffraction grating. On interacting the material, it will get diffract towards a glass prism, the prism will converge these beams on a single point on the screen (As shown in the picture).

As we can see from figure that there are several rays passing but we will consider just ray 1 & 2 in our further processing. From diffraction grating, ray 1 is covering more distance than the ray 2 in order to reach prism. To find the path difference lets draw a perpendicular from ray 2 on ray 1. Here ‘Ar’ represents the bath difference between these 2 waves. From the triangle ABr we have,

Where,

Ar = AB sinθ —————(1)

Here Ar is the path difference & AB = d. Therefore, we can rewrite equation (1)

a Path difference = d sinθ ……………………….. (2)

But for constructive interference, we have

Path difference = mλ …………………………… (3)

Where, m = 0, 1, 2, 3, 4, ………

Comparing equation (2) and (3), we get

d Sinθ = mλ

If we know the value of ‘θ’ and ‘d’ then above equation can be used to calculate the wavelength of light.

**Q.8) Describe the construction and working of Michelson’s interferometer. How one can determine the wavelength of light used by this instrument?**

**Answer:**

**Michelson’s Interferometer:**

Michelson Interferometer is an optical instrument. This instrument is used to study the interference of light waves and in order to find its wavelength.**Diagram:**

**Working:**

Its working principle is based on the division of amplitude. This division usually takes place by partial reflection and transmission of light at the boundary of two mediums. The interferometer consists of two plane mirrors M_{1} which is movable and M_{2} which is fixed and two lass plates A and B arranged according to figure.

Plate A is made light silver on its back so that it may reflect half of the incident light on it and let another half to transmit. A beam of light from source ‘S’ falls on plate ‘A’ and will split into two parts.

The first part will reflect from plate A and directed towards the mirror M_{1}. After reflecting from M1 it will pass through plate A and through a telescope it will enter the eye. The second part will transmit from plate A and will direct towards the mirror M2 which after reflection will enter to the eye as well. Plate B will serve as a compensator for the second beam to make a path of both beams to be equal.**Calculation of wavelength:**

Consider there is ‘m’ number of fringes. Therefore, when M1 is moved backwards through a distance of λ/4 each time, then the total covered distance ‘P’ is given by,

Above equation helps us to calculate wavelength of light used if we know the value of P and m.

**Q.9) What is meant by plane polarized light? How does this phenomenon decide that light waves are transverse in nature.**

**Polarization:**

“Polarization is the process by which the electric and magnetic vibrations of light waves are restricted to a single plane of vibration”

**Plane Polarized Light:**

The process of forcing or confining of light waves to vibrate in one plane is known as plane polarization and such polarized light is called plane-polarized light. Polarization property cannot be an exhibit by longitudinal or sound waves. It is a property which only transverse waves can possess.

**Experiment with Transverse Waves:**

Consider a wooden box having a slot in it. Let a transverse wave on a string passes through that slot. When the slot will be parallel to the direction of vibration of passing wave then the wave will go through the slot undisturbed. But if we place another slot such that the two slots are perpendicular to each other, then there will be no passing of any wave.

But if we place a spring instead of the string then the wave will pass through the slots regardless of the orientation of the slots. Therefore, it is clear that longitudinal waves do not show the polarization property but only transverse waves does.

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