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NCERT solutions for class 12 Physics chapter 10 Wave Optics - Students appearing in the Class 12 board exams must go through this Solutions for Class 12. These NCERT solutions for Class 12 Physics chapter 10 Wave Optics gives you an insight into how to deal with the questions asked in the NCERT books for Class 12 Physics. NCERT Class 12 Physics solutions chapter 10 Wave Optics comprise of the step by step solutions in an easy-to-understand language. The NCERT solutions for Class 12 Physics chapter 10 Wave Optics are helpful in academics as well as in competitive exams. Read further to know the Class 12 Physics Chapter 10 NCERT solutions.
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NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics: Exercises Solution
Answer:
Given a monochromatic light has a wavelength of .
And as we know the speed of the light in air is, .
Refractive index of water,
Therefore we have now,
The ray is incident on the medium, and then it gets reflected back to the same medium therefore, the wavelength, speed, and frequency will be the same as that of the incident ray.
We have the frequency of light is given by the relation,
Therefore the speed, frequency, and the wavelength of the reflected light are , , and respectively.
Answer:
Given,
The wavelength of the monochromatic light
Refractive index of the water
b)In the case of refraction, speed and wavelength will change as the medium is changing however, the frequency will remain the same because it doesn't change when the medium is changed,
so, frequency
Speed of rays: speed of rays in water with refractive index 1.33 is
Now, the Wavelength of light
Hence the wavelength of light is 444.01nm, frequency is and speed is .
Q10.2 (a) What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
Answer:
The shape of wavefront when light is diverging from a point source is Spherical since light travels in all direction.
Q10.2 (b) What is the shape of the wavefront in each of the following cases:
(b) Light emerging out of a convex lens when a point source is placed at its focus.
Answer:
The shape of light emerging out of a convex lens when a point source is placed at its focus is Parallel .when light rays come from infinity parallelly, they intersect at focus of convex lens and hence when light is emerging from the focus, the rays will get parallel to each other after coming out of the convex lens, because path of light rays are reversible.
Q10.2 (c) What is the shape of the wavefront in each of the following cases:
(c) The portion of the wavefront of light from a distant star intercepted by the Earth
Answer:
The portion of the wavefront of the light from the distant star which is seen from earth is plane since a small area of a large sphere will nearly look like a plane.
Answer:
Given,
Refractive index of the glass = 1.5
Speed of light in vaccum
Now,
As we know,
Refractive index of a medium
where is the speed of light in that medium.
so from here,
Hence the speed of light in water is
Answer:
No, the speed of light in glass is not independent of the colour. The colour of the light does influence the refractive index and speed of light in the medium. The refractive index of the violet light is greater than the refractive index of red light and hence red component of the white light travels faster in the glass than the red component of the light.
as
the more the refractive index, the lesser the speed.
Answer:
Given,
Distance between screen and slit
Distance between slits
Distance between central and fourth bright fringe
Now,
as we know, the distance between two fringes in a constructive interference is given by
where order of fringe (which is 4 here) and is the wavelength of light we are using.
so from here,
Hence wavelength os the light is 600nm
Q10.5 In Young’s double-slit experiment using monochromatic light of wavelength , the intensity of light at a point on the screen where path difference is , is K units. What is the intensity of light at a point where path difference is ?
Answer:
Given, in youngs double-slit experiment.
the wavelength of monochromatic light =
The intensity of light when the path difference is = K
Now,
As we know,
The phase difference is given by
also
Total Intensity
Let
Now, when path difference is
the phase difference angle
so,
Now, when path difference is
Intensity of light
Now comparing intensity at both cases
Hence intensity will reduce to one-fourth of initial when path difference changes from to .
Q10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
Answer:
Given,
The wavelength of one light beam :
The wavelength of another Light beam
Let, the distance between the two-slit be and distance between slit and screen is
Now,
As we know, the distance of nth bright fringe from central maxima is given by
so for 3rd fringe,
Hence distance of 3rd fringe from central maxima is . Here value D and d are not given in the question.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Answer:
Let the nth bright fringe due to wavelength, and (n − 1) th bright fringe due to wavelength coincide on the screen. We can equate the conditions for bright fringes as:
Hence, the least distance from the central maximum can be obtained by the relation:
Hence we can find out X if D and d are given.
Answer:
Given
The angular width of the fringe when the medium is air
The distance of the screen from the slit
The wavelength of light we are using
Refractive index of water
let angular width of fringe when the medium is water
Now, as we know the angular width is given by
so,
From here
Hence angular width of the fringe in the water is .
Q10.8 What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Answer:
Given,
Refractive index of glass
Now as we know,
where is the polarizing angle, also called the Brewster angle.and is the refractive index.
so from here
Hence Brewster angle is .
Answer:
Given,
Wavelength of light
Speed of light
Now,
Wavelength and frequency will be the same when the ray is reflected.
Frequency of reflected light
Hence wavelength and frequency of light is and respectively.
Now,
as per the law of reflection, angle of incidence is always equal to angle of reflection .
Now, when the reflected ray is perpendicular with incidence ray,
Hence the angle of incidence is 45 for this condition.
Answer:
Given
Aperture
Wavelength of light
Now,
Distance for which ray optics is a good approximation also called Fresnel's distance:
Hence distance for which ray optics is a good approximation is 40m.
Answer:
Given,
wavelength Hα line emitted by hydrogen:
star is red-shifted by
let velocity of the star be
Now,
as we know,
from here
Hence speed at which star is receding away is
Answer:
According to corpuscular theory, when corpuscle of the light goes from rare medium to denser medium, the component of their velocity along the surface of the interface remains the same.
So we can write
As ,
That is light should be faster in the dense medium than in rare medium.this is the opposite of what we see experimentally.
Huygens wave theory predicts that light is faster in a rare medium which matches with our experiments observation.
Answer:
Let an object M is placed in front of a plane mirror AB at a distance r .
A circle is drawn from the centre, such that it just touches the plane mirror at point P. according to the Huygens’ principle, AB is the wavefront of the incident light .
If the mirror is absent then a similar wavefront A'B' would form behind M at a distance r.
A'B' can be considered as a virtual reflected ray for the plane mirror. Hence a point object placed in front of the plane mirror produces an image at the same distance as the object.
(iv) wavelength.
(v) intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum depend?
Answer:
The speed of light in a vacuum is constant and independent of anything according to Einstein's theory of relativity.
Question:
(b) the speed of light in a medium (say, glass or water), depend?
Answer:
The speed of light in any medium depends upon the wavelength of the light and does not depends on the nature of the source, direction of propagation, the motion of the source and/or observer, and intensity of the wave.
Answer:
The sound wave requires a medium for propagation.so, even though both given situation may relate to the same relative motion, they are not identical physically since, the motion of the observer, relative to the medium is different in two situations. Hence, we cannot expect the Doppler formula to be identical in both given cases.
When light waves are in a vacuum, there is clearly nothing to distinguish between two cases.
for light propagation in a medium, two situations are not identical for the same reason as in the case of sound waves.
Answer:
Given,
Wavelength of light = 600nm
Angular fringe width
Hence spacing required between the two slits is .
Q10.17 (a) Answer the following questions:
In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
Answer:
As we know,
width of the central diffraction band is given by
where d is the width of the slit.
So when we double the width of the slit, the size of the central diffraction band reduces to half of its value. But, the light amplitude becomes double, which increase the intensity 4 times.
10.17 (b) Answer the following questions:
Answer:
When we have a width in the order of , the intensity of interference fringes in Young's double-slit experiment is modified by the diffraction pattern of each slit.
Q10.17 (c) Answer the following questions:
Answer:
A bright spot is seen at the centre of the shadow of the obstacle because wave diffracted from the edge of a circular obstacle interfere constructively at the centre of the shadow producing the bright spot.
Q10.17 (d) Answer the following questions:
Answer:
The size of obstacle should be comparable to the wavelength for diffraction of waves by obstacles, through a large scale.
This comes from
This implies
it means the light goes almost unbent and hence student are unable to see each other.
Q10.17 (e) Answer the following questions:
Answer:
Typical size of the obstacle is much larger than the wavelength of light. Hence the diffraction effect is negligibly small. thus the assumption that light travels in a straight line can be safely used in day to day life.
Answer:
Given,
Distance between two towers = 40km
size of aperture =
Now,
As we know
Fresnel's distance is equal to half of the distance between towers
Also from the formula:
Hence this is the required longest wavelength of the radio wave, which can be sent in between the towers without considerable diffraction effect.
Q10.19 A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Answer:
Given
The distance of the screen from the slit,
The distance of the first minimum
The wavelength of the light
Now,
As we know,
Hence, the width of the slit is 0.2 mm.
Answer:
When a low flying aircraft passes overhead, we notice slight shaking in pictures of the TV. This is because aircraft interferes with signals and reflects it. So the shaking we see is the interference of direct signal and reflected signal.
Q10.20 (b) Answer the following questions:
Answer:
The superposition principle comes from the linear character of the differential equation of wavemotion.that is if and are the solution of any wave equation, then linear combination of and is also the solution of the wave equation.
Answer:
Let the width of the slit be divided into n equal parts so that
Now,
At this angle, each slit will make the first diffraction minimum. therefore the resultant intensity for all the slits will be zero at the angle of .
Wave optics class 12 exercise solutions hold great significance for board exams, as well as competitive exams like JEE and NEET. This chapter's questions range from moderately challenging to complex, offering students the opportunity to earn substantial marks. By thoroughly understanding concepts related to interference, diffraction, and polarization, and practising with these solutions, students can confidently tackle questions and secure good marks in both board and competitive exams.
Chapter wise NCERT Solutions for class 12 physics
Path difference (Δ):
T.D. (Time Difference):
Resultant Amplitude:
Resultant intensity:
Path difference:
Where x be the position of point P from central maxima
For maxima at P:
For minima at P:
Brewster’s law:
Resolving power of Microscope:
Resolving power of Telescope:
The topics covered in the Wave Optics Class 12 NCERT Physics are Huygens Principle, Explanation of Refraction and Reflection using Huygens Principle, Addition of Waves, Interference, Defraction, Polarisation and related subtopics. One important experiment of Wave Optics is Young's Double Slit Experiment.
Previously, in the CBSE board exam, 11-mark questions were asked from the unit optics which includes chapters 9 Ray Optics and Wave Optics of NCERT Class 12 Physics.
For competitive exams also the Wave Optics NCERT solutions are important. Questions from the topic of young's double-slit experiment are frequently asked in competitive exams like NEET and JEE Main.
NCERT solutions for Class 12 Physics Chapter 10 wave optics are important to score well in exams.
Comprehensive Explanations: These wave optics class 12 exercise solutions offer comprehensive explanations, aiding students in understanding intricate wave optics concepts.
Problem-Solving Aid: Students can utilize these ncert class 12 physics chapter 10 pdf to practice and enhance their problem-solving skills in wave optics.
Clarity in Language: The solutions are presented in straightforward language, ensuring students can easily grasp and apply the principles.
Exam Preparation: These wave optics ncert solutions are a valuable resource for preparing for both board exams and competitive exams like JEE and NEET.
Performance Enhancement: By leveraging these solutions, students can improve their overall performance in physics.
Accessible for All: These wave optics class 12 solutions are freely accessible, making them available to all students seeking assistance with wave optics.
Firstly go through the complete NCERT syllabus for Class 12 Physics and check all the topics. Through this, you will get a clear understanding of what to study.
Afterwards, try to solve the questions on your own before attempting the solutions.
If you still have any doubts then check the Wave Optics Class 12 NCERT Solutions.
Along with the Class 12 Physics Chapter 10 NCERT solutions, solve the previous year’s question papers and sample papers too.
NCERT solutions subject-wise
Yes, NCERT books are enough to prepare for the board exams, but you can refer to other reference books and sample papers as well. Try to cover all the concepts based on the NCERT syllabus. To get a good score in the CBSE board exam understand all the topics in the NCERT book and solve all the questions of NCERT exercise. Additionally students can refer NCERT exemplar problems and CBSE previous year question papers.
Yes, NCERT solutions will be helpful for competitive exams as well. Solving NCERT problems will give a better idea of concepts studied in a chapter and this in turn helps in competitive exams like JEE Main ans NEET.
Questions will be based on NCERT topics. But may be an application level question. The questions maynot be directly from the NCERT Questions but are related to the NCERT syllabus.
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