Physics Class 11 Notes Physical Optics Chapter 9 for kpk

Physics Class 11 Notes Physical Optics Chapter 9 for kpk

Physics Class 11 Notes Physical Optics Chapter 8 for kpk, fbise, punjab, and sindh boads.

Physical Optics Physics 1st Years Notes

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 ‘iP’ 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
n1 sin iP = n2 sin r ……..………………………….. (1)
Where n1 and n2 are absolute refractive indexes of medium 1 and 2, respectively. From figure, we can write
iP + 900 + r = 1800
⇒ r = 1800–900 — iP = 900 — iP
Putting this value in equation (1), we get
n1 sin iP = n2 sin (900 — iP) = n2 cos iP

Physics Class 11 Notes Physical Optics Chapter 8 for kpk
Physics Class 11 Notes Physical Optics Chapter 8 for kpk

Read more: Physics Class 11 Notes Waves Chapter 8 for kpk 2021

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

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?

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?

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?

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?

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?

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.

Read more: Physics Class 11 Notes 2021 Oscillation Chapter 7 for kpk

Physics Class 11 2021 comprehension question Chapter 9


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 × 108 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?


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.

Physics Class 11 Notes Physical Optics Chapter 8 for kpk 2021
Physics Class 11 Notes Physical Optics Chapter 8 for kpk 2021

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.

Physics Class 11 Notes Physical Optics Chapter 8  2021
Physics Class 11 Notes Physical Optics Chapter 8  2021

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 mth bright fringes we have

Physics Class 11 Notes Physical Optics Chapter 8 for kpk fbise and sindh
Physics Class 11 Notes Physical Optics Chapter 8 for kpk fbise and sindh

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.


Physics Class 11 Notes Physical Optics Chapter 8 for kpk pdf
Physics Class 11 Notes Physical Optics Chapter 8 for kpk pdf

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.

“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.

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