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NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

Edited By Vishal kumar | Updated on Sep 11, 2023 01:26 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Physics Chapter 10 – Download Free PDF

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

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NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics: Exercises Solution

Q10.1 (a) Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected. Refractive index of water is 1.33.

Answer:

Given a monochromatic light has a wavelength of 589 nm\ or\ 589\times 10^{-9}m .

And as we know the speed of the light in air is, c = 3\times 10^{8} m/s .

Refractive index of water, \mu = 1.33

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,

v = \frac{c}{\lambda}

= \frac{3\times 10^8 m/s}{589\times 10^{-9}}

= 5.09\times 10^{14}Hertz = 5.09\times 10^{14}Hertz

Therefore the speed, frequency, and the wavelength of the reflected light are 3\times 10^{8} m/s , 5.09\times 10^{14}Hertz , and 589 nm respectively.

Q10.1 (b) Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (b) refracted light? Refractive index of water is 1.33.

Answer:

Given,

The wavelength of the monochromatic light \lambda =589nm=589*10^{-9}m

Refractive index of the water \mu_{water}=1.33

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 f=5.09 * 10^{14}Hz

Speed of rays: speed of rays in water with refractive index 1.33 is

v=\frac{c}{\mu _{water}}=\frac{3*10^8}{1.33}=2.26*10^8m/s

Now, the Wavelength of light

\lambda =\frac{v}{f}=\frac{2.26*10^8}{5.09*10^{14}}=444.01*10^{-9}m

Hence the wavelength of light is 444.01nm, frequency is 5.09 * 10^{14}Hz and speed is 2.26*10^8m/s .

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.

Q10.3 (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 \times 10^8 m s^{-1} )

Answer:

Given,

Refractive index of the glass \mu _{glass} = 1.5

Speed of light in vaccum c=3*10^8m/s

Now,

As we know,

Refractive index of a medium

\mu _{medium }=\frac{c}{v_{medium}} where {v_{medium}} is the speed of light in that medium.

so from here,

{v_{glass}}=\frac{c}{\mu _{glass}}=\frac{3*10^8}{1.5}=2*10^8m/s

Hence the speed of light in water is 2*10^8m/s

Q10.3 (b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

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

v=\frac{c}{\mu }

the more the refractive index, the lesser the speed.

Q10.4 In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Answer:

Given,

Distance between screen and slit D=1.4m

Distance between slits d=0.28mm=0.28*10^{-3}m

Distance between central and fourth bright fringe u=1.2cm=1.2*10^{-2}m

Now,

as we know, the distance between two fringes in a constructive interference is given by

u=n\lambda \frac{D}{d}

where n= order of fringe (which is 4 here) and \lambda is the wavelength of light we are using.

so from here,

\lambda = \frac{ud}{nD}=\frac{1.2*10^{-2}*0.28*10^{-3}}{4*1.4}=6*10^{-7}m

Hence wavelength os the light is 600nm

Q10.5 In Young’s double-slit experiment using monochromatic light of wavelength \lambda , the intensity of light at a point on the screen where path difference is \lambda , is K units. What is the intensity of light at a point where path difference is \frac{\lambda}{3} ?

Answer:

Given, in youngs double-slit experiment.

the wavelength of monochromatic light = \lambda

The intensity of light when the path difference is \lambda = K

Now,

As we know,

The phase difference \phi is given by

\phi =\frac{2\pi }{\lambda }(PathDiffernce)

also

Total Intensity

I=I_1+I_2+2\sqrt{I_1I_2}cos\phi

Let I_1=I_2=I_0

Now, when path difference is \lambda

the phase difference angle

\phi=\frac{2\pi }{\lambda}*\lambda=2\pi

so,

I_0+I_0+2\sqrt{I_0I_0}cos2\pi=K

K=4I_0

Now, when path difference is \frac{\lambda }{3}

\phi=\frac{2\pi }{\lambda}*\frac{\lambda}{3}=\frac{2\pi}{3}

Intensity of light

K'=I_0+I_0+2\sqrt{I_0I_0}cos\frac{2\pi}{3}

K'=I_0

Now comparing intensity at both cases

\frac{K'}{K}=\frac{I_0}{4I_0}=\frac{1}{4}

K'=\frac{K}{4}

Hence intensity will reduce to one-fourth of initial when path difference changes from \lambda to \frac{\lambda}{3} .

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 :

\lambda_1=650nm=650*10^{-9}m

The wavelength of another Light beam

\lambda_2=520nm=520*10^{-9}m

Let, the distance between the two-slit be d and distance between slit and screen is D

Now,

As we know, the distance x of nth bright fringe from central maxima is given by

x=n\lambda \frac{D}{d}

so for 3rd fringe,

n=3

x=n\lambda_1 \frac{D}{d}=3*650*10^{-9}*\frac{D}{d}=1950\frac{D}{d}nm

Hence distance of 3rd fringe from central maxima is 1950\frac{D}{d}nm . Here value D and d are not given in the question.

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.

(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, \lambda_2 and (n − 1) th bright fringe due to wavelength \lambda_1 coincide on the screen. We can equate the conditions for bright fringes as:

n\lambda_2=(n-1)\lambda _1

520n=650n-650

650=130n

n=5

Hence, the least distance from the central maximum can be obtained by the relation:

X=n\frac{\lambda_2D}{d}=5*\frac{520*10^{-9}*D}{d}

Hence we can find out X if D and d are given.

Q10.7 In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.

Answer:

Given

The angular width of the fringe when the medium is air

\theta _{air}=0.2^0

The distance of the screen from the slit D = 1m

The wavelength of light we are using \lambda=600nm=600*10^{-9}m

Refractive index of water \mu_{water}=4/3

let angular width of fringe when the medium is water \theta _{water}

Now, as we know the angular width is given by

\theta =\frac{\lambda }{d}

so,

d=\frac{\lambda _{air}}{\theta _{air}}=\frac{\lambda _{water}}{\theta _{water}}

d=\frac{\lambda _{air}}{\lambda _{water}}=\frac{\theta _{air}}{\theta _{water}}=\mu

From here

\theta _{water}=\frac{\theta _{air}}{\mu _{water}}=\frac{3}{4}0.2^0=0.15^0

Hence angular width of the fringe in the water is 0.15^0 .

Q10.8 What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)

Answer:

Given,

Refractive index of glass \mu_{glass}=1.5

Now as we know,

tan\theta =\mu

where \theta is the polarizing angle, also called the Brewster angle.and \mu is the refractive index.

so from here

\theta =tan^{-1}(\mu _{glass})=tan^{-1}1.5=56.31^0

Hence Brewster angle is 56.31^0 .

Q10.9 Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Answer:

Given,

Wavelength of light \lambda =5000\AA =5000*10^{-10}m

Speed of light c=3*10^8

Now,

Wavelength and frequency will be the same when the ray is reflected.

Frequency of reflected light

f=\frac{c}{\lambda }=\frac{3*10^{8}}{5000*10^{-10}}=6*10^{14}Hz

Hence wavelength and frequency of light is 5000*10^{-10}m and 6*10^{14}Hz respectively.

Now,

as per the law of reflection, angle of incidence i is always equal to angle of reflection r .

i=r

Now, when the reflected ray is perpendicular with incidence ray,

i+r=90

i+i=90

2i=90

i=45

Hence the angle of incidence is 45 for this condition.

Q10.10 Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.

Answer:

Given

Aperture a=4mm=4*10^{-3}m

Wavelength of light \lambda =400nm=400*10^{-9}m

Now,

Distance for which ray optics is a good approximation also called Fresnel's distance:

Z_f=\frac{a^2}{\lambda }=\frac{(4*10^{-3})^2}{400*10^{-9}}=40m

Hence distance for which ray optics is a good approximation is 40m.


NCERT solutions for Class 12 Pysics chapter 10 Wave Optics additional exercises:

Q10.11 The 6563 Å Hα line emitted by hydrogen in a star is found to be red-shifted by 15 Å. Estimate the speed with which the star is receding from the Earth.

Answer:

Given,

wavelength Hα line emitted by hydrogen:

\lambda=6563*10^{-10}m

star is red-shifted by

\lambda'-\lambda=15*10^{-10}m

let velocity of the star be v

Now,

as we know,

\lambda'-\lambda=\frac{v}{c}*\lambda

from here

\lambda'-\lambda\frac{1}{{c}*\lambda }={v}

v=\frac{3*10^8*15*10^{-10}}{6563*10^{-10}}=6.87*10^5m/s

Hence speed at which star is receding away is 6.87*10^5m/s

Q10.12 Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?

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

v_1sini=v_2sinr

\frac{v_2}{v_1}=\frac{sini}{sinr}=\mu

As \mu>1 ,

v_2>v_1

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.

Q10.13 You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.

Answer:

Let an object M is placed in front of a plane mirror AB at a distance r .


1649156499396

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.

1649156518145

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.

Q10.14 (a) Let us list some of the factors, which could possibly influence the
speed of wave propagation:
(i) nature of the source.
ii) direction of propagation.
(iii) motion of the source and/or observer.

(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:

Q10.14 (b) Let us list some of the factors, which could possibly influence the speed of wave propagation:
(i) nature of the source.
(ii) direction of propagation.
(iii) motion of the source and/or observer.
(iv) wavelength.
(v) intensity of the wave.
On which of these factors, if any, does

(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.

Q10.15 For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?

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.

Q10.16 In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?

Answer:

Given,

Wavelength of light \lambda = 600nm

Angular fringe width

B_{\theta}=\frac{\lambda}{d}

d=\frac{\lambda}{B_{\theta}}=\frac{600*10^{-9}*180}{0.1*\pi }=3.44*10^{-4}m

Hence spacing required between the two slits is 3.44*10^{-4}m .

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

2D\frac{\lambda}{d}

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:

In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?

Answer:

When we have a width in the order of \lambda , 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:

When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?

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:

Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.

Answer:

The size of obstacle should be comparable to the wavelength for diffraction of waves by obstacles, through a large scale.

This comes from

sin\theta =\frac{\lambda}{a}

sin\theta =\frac{10^{-7}}{10}=10^{-8}

This implies \theta\rightarrow 0

it means the light goes almost unbent and hence student are unable to see each other.

Q10.17 (e) Answer the following questions:

Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?

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.

Q10.18 Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?

Answer:

Given,

Distance between two towers = 40km

size of aperture = a=50m

Now,

As we know

Fresnel's distance is equal to half of the distance between towers

Z_f=\frac{40}{2}=20km


Also from the formula:

Z_f=\frac{a^2}{\lambda}=20m

\lambda=\frac{a^2}{Z_f}=\frac{50^2}{20*10^3}=12.5cm

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, D=1m

The distance of the first minimum X_1=2.5mm=10^{-3}=2.5*10^{-3}mm

The wavelength of the light \lambda=500nm=500*10^{-9}m

Now,

As we know,

X_n=n\frac{\lambda D}{d}

d=n\frac{\lambda D}{X_n}=1*\frac{500*10^{-9}*1}{2.5*10^{-3}}=2*10^{-4}m=0.2mm

Hence, the width of the slit is 0.2 mm.

Q10.20 (a) Answer the following questions:
When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.

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:

As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?

Answer:

The superposition principle comes from the linear character of the differential equation of wavemotion.that is if x_1 and x_2 are the solution of any wave equation, then linear combination of x_1 and x_2 is also the solution of the wave equation.

Q10.21 In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of \frac{n\lambda}{a} . Justify this by suitably dividing the slit to bring out the cancellation.

Answer:

Let the width of the slit b be divided into n equal parts so that

b'=\frac{b}{n}

b=b'n

Now,

\theta=\frac{n\lambda}{b}=\frac{n\lambda}{b'n}=\frac{\lambda}{b'}

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 \frac{n\lambda}{b} .

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

Wave Optics Class 12 Solutions: Important Formulas and Diagrams

  • Huygens' Principle

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Path difference (Δ): 1644993812368


T.D. (Time Difference): 1644993809828


Resultant Amplitude: 1644993810103

Resultant intensity: 1644993815082

  • Young’s Double Slit Experiment (YDSE)

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1694418885717

Path difference: 1644993822987

Where x be the position of point P from central maxima

For maxima at P: 1644993822757 1644993817657

For minima at P: 1644993820110 1644993817424

  • Brewster’s law: 1644993823231

  • Resolving power of Microscope: 1644993824253

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Resolving power of Telescope: 1644993811723

Also Check NCERT Books and NCERT Syllabus here:

Wave Optics NCERT Topics

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.

Importance of NCERT solutions for class 12 physics chapter 10 wave optics:

  • 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.

Key features of Wave Optics Class 12 NCERT Solutions

  1. Comprehensive Explanations: These wave optics class 12 exercise solutions offer comprehensive explanations, aiding students in understanding intricate wave optics concepts.

  2. 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.

  3. Clarity in Language: The solutions are presented in straightforward language, ensuring students can easily grasp and apply the principles.

  4. Exam Preparation: These wave optics ncert solutions are a valuable resource for preparing for both board exams and competitive exams like JEE and NEET.

  5. Performance Enhancement: By leveraging these solutions, students can improve their overall performance in physics.

  6. Accessible for All: These wave optics class 12 solutions are freely accessible, making them available to all students seeking assistance with wave optics.

How to use NCERT solutions for class 12 Physics chapter 10 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

Frequently Asked Question (FAQs)

1. Are NCERT books for Class 12 Physics enough to prepare for CBSE Board exams?

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.

2. Will the NCERT solutions be helpful for competitive exams as well?

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.

3. Will the questions in the CBSE board exams be directly asked from the NCERT books?

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.

4. How can I quickly solve the wave optics-based problems in Chapter 10 of wave optics class 12 ncert?

To quickly solve ncert wave optics problems in ch 10 physics class 12, make sure you have a clear understanding of the fundamental concepts and practice solving problems. Read the problem statement carefully and use diagrams to visualize the problem. Review your work and seek help if needed.

5. How wave optics ncert solutions are important for NEET?

wave optics class 12 ncert solutions is important for NEET as the exam covers a wide range of topics from physics and wave optics is one of the important topics that is covered. Concepts and problem-solving skills learned in this chapter will be useful for understanding and solving questions related to wave optics that may appear on the NEET exam.



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Questions related to CBSE Class 12th

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hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Hi there,

Yes you are obviously eligible to participate  in the jee mains and advance examination in 2025. Since it was your first attempt in the year 2024, you can certainly take a one year gap , that is a drop year and can give the exam in the next academic year.


For appearing in the exam in 2025, you only have to give a gap certificate to confirm your participation in the upcoming exam. You can further read more details regarding the gap year and the documents required to further participate in the examination.


Hope this resolves your query.

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.


Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.


Hope this resolves your query.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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