Course Content
Class 12th Physics Chapter 1 Electric Charges And Fields
NCERT Solutions for Chapter 1 for Class 12 Physics talks about factors and topics related to Electric Charge and Fields. Some of the prominent topics covered in the Chapter are Conductors And Insulators, Charging By Induction, Basic Properties Of Electric Charge, Coulomb’s Law, etc.
Class 12th Physics Chapter 2 Electrostatic Potential And Capacitance
NCERT Solutions for Chapter 2 for Class 12 Physics talks about Electrostatic Potential in detail. The chapter revolves around relevant topics such as Potential Due To A Point Charge, Potential Due To An Electric Dipole, Potential Due To A System Of Charges, Equipotential Surfaces, etc.
Class 12th Physics Chapter 3 Current Electricity
NCERT Solutions for Chapter 3 for Class 12 Physics is an important chapter from weightage point of view. This chapter is very scoring and can be a very interesting chapter for students. The chapter talks about Electric Current, Ohm’s Law, Drift Of Electrons And The Origin Of Resistivity, Limitations Of Ohm’s Law, Resistivity Of Various Materials, etc.
Class 12th Physics Chapter 4 Moving Charges And Magnetism
Chapter 4 for Class 12 Physics can be a very scoring chapter for Class 12 students. However, it is equally complex to understand. It talks about Magnetic Force, Motion In A Magnetic Field, Motion In Combined Electric And Magnetic Fields, etc.
Class 12th Physics Chapter 5 Magnetism And Matter
NCERT Solutions for Chapter 5 for Class 12 Physics talk about topics related to Magnetism and Matter such as the Bar Magnet, Magnetic field lines, The electrostatic analogue, Magnetism And Gauss’s Law, The Earth’s Magnetism, Magnetic declination and dip, Magnetic Properties Of Materials, Diamagnetism, Paramagnetism, Ferromagnetism.
Class 12th Physics Chapter 6 Electromagnetic Induction
A very interesting and Chapter, this one in particular talks about The Experiments Of Faraday And Henry, Magnetic Flux, Faraday’s Law Of Induction, Lenz’s Law and Conservation Of Energy.
Class 12th Physics Chapter 7 Alternating Current
NCERT Solutions for Chapter 7 for Class 12 Physics talk about Ac Voltage Applied To A Resistor, Phasor-diagram solution, Analytical solution, Resonance, Oscillations, and Transformers.
Class 12th Physics Chapter 8 Electromagnetic Waves
In this chapter students will learn about Displacement Current, Electromagnetic Waves, Sources of electromagnetic waves and other such topics related to Electromagnetic Waves.
Class 12th Physics Chapter 9 Ray Optics and Optical Instruments
NCERT Solutions for Chapter 9 for Class 12 Physics talks about Reflection Of Light By Spherical Mirrors, Refraction, Total Internal Reflection and other topics related to Ray Optics.
Class 12th Physics Chapter 10 Wave Optics
NCERT Solutions for Chapter 10 for Class 12 Physics talk about Huygens Principle, Refraction and Reflection Of Plane Waves Using Huygens Principle, Refraction Of A Plane Wave, Refraction At A Rarer Medium etc.
Class 12th Physics Chapter 11 Dual Nature Of Radiation And Matter
Chapter 11 for Class 12 Physics is a scoring chapter. If learnt the concepts thoroughly, it can help students to score high marks in their Physics exam. This chapter talks about Electron Emission, Photoelectric Effect, Hertz’s observations, Hallwachs’ and Lenard’s observations, etc.
Class 12th Physics Chapter 12 Atoms
Chapter 12 for Class 12 Physics is a very interesting yet very confusing chapter for the students. It talks about topics: Alpha-Particle Scattering and Rutherford’s Nuclear Model Of Atom, Alpha-particle trajectory, Electron orbits, Atomic Spectra etc.
Class 12th Physics Chapter 13 Nuclei
NCERT Solutions for Chapter 13 for Class 12 include all answers to the questions included in the textbook. Additionally, the solutions explain topics such as Atomic Masses And Composition Of Nucleus, Size Of The Nucleus, Mass-energy And Nuclear Binding Energy, etc in detail.
Class 12th Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuits
Chapter 14 for Class 12 Physics Is a very scoring and interesting chapter and revolves around topics such as Intrinsic Semiconductor, Extrinsic Semiconductor, P-n Junction, Semiconductor Diode, Application Of Junction Diode As A Rectifier etc.
Class 12th Physics Chapter 15 Communication Systems
NCERT Solutions for Chapter 15 for Class 12 Physics talks about topics related to the communication systems. The important topics are Elements Of A Communication System, Basic Terminology Used In Electronic Communication Systems, etc. This is a very interesting chapter in the Class 12 Physics textbook.
Frequently asked questions (FAQs) Related to Course
Here are some frequently asked questions (FAQs) that could be relevant for your Class 12 Physics Online Course: 1. What is the Class 12 Physics Online Course offered by Free-Education (Wisdom Academy)? Our Class 12 Physics Online Course is a comprehensive learning program designed to help students excel in their CBSE Board exams. It includes chapter-wise notes, important questions, solved and unsolved previous year question papers, video explanations, interactive classes, and more. 2. Who is the instructor for this course? The course is curated and taught by Er. Mohd Sharif, an experienced educator with over 13 years of teaching Physics. He is also the founder of Wisdom Education Academy & Free-Education.In. 3. How can I access the course materials? Once you enroll in the course, you will gain access to our online learning platform. All course materials, including notes, videos, and question papers, can be accessed from there at any time. 4. Are the notes and resources provided in downloadable formats? Yes, our chapter-wise notes and other resources are available in downloadable PDF formats. This allows you to study even when you're offline. 5. How can I ask questions or clear doubts during the course? We offer doubt clearance through online classes, where you can interact with the instructor directly and get your queries resolved. You can also use our platform's messaging system to communicate with the instructor. 6. Are there regular tests to review my preparation? Absolutely, we believe in consistent evaluation. The course includes regular review tests to gauge your understanding and track your progress. 7. Can I access the course on mobile devices? Yes, our online learning platform is mobile-friendly, allowing you to access the course materials on various devices like smartphones and tablets. 8. Are there any prerequisites for enrolling in this course? This course is designed for Class 12 students studying Physics. It's tailored to cover all topics and concepts within the Class 12 Physics curriculum. 9. How can I enroll in the Class 12 Physics Online Course? Enrolling is simple. Visit our website, navigate to the Class 12 Physics Course, and follow the enrollment instructions. 10. Is there a refund policy in case I'm not satisfied with the course? Yes, we have a refund policy in place. If you're not satisfied with the course within the first 7 days, you can request a refund according to our refund policy. 11. Can I interact with other students in the course? Yes, we encourage collaborative learning. Our platform may include discussion forums or other means for students to interact and share insights. 12. Are there any additional resources provided apart from the course materials? In addition to the course materials, we may provide supplementary resources like articles, study tips, and external references to enhance your learning experience. For any other questions or clarifications, feel free to reach out to our support team at (8750387081).
Class 12 Physics Online Course: Notes, Important Questions, and Previous Year Papers
About Lesson

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Question 1. ( ncert solution ray optics )
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
The object is kept between ƒ and C. So the image should be real, inverted and beyond C. To locate the sharp image, screen should be placed at the position of image.

So, the image is inverted and magnified. Thus in order to locate the sharp image, the screen should be kept 54 cm in front of concave mirror and image on the screen will be observed real, inverted and magnified. If the candle is moved closer to the mirror, the real image will move away from the mirror, hence screen has to be shifted away from the mirror to locate the sharp image.

Question 2. ( ncert solution ray optics )
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
A convex mirror always form virtual image, which is erect and small in size of an object kept in front of it. Focal length of convex mirror ƒ = + 15 cm Object distance u = – 12 cm Using mirror formula

Therefore, image is virtual, formed at 6.67 cm at the back of the mirror.

It shows the image is erect, small in size and virtual. When the needle is moved farther from mirror the image also move towards focus and decreasing in size. As u approaches v approaches focus but never beyond ƒ.

Shift = Real thickness – Apparent of thickness

Question 18.
Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we’see’a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake.
(d) Does the apparent depth of a tank of water change if viewed obliquely? if so, does the apparent depth increase of decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. is this fact of some use to a diamond cutter?
(a) In this situation when rays are convergent behind the mirror, both plane mirror and convex mirror can form real images of virtual objects.

index of diamond is much greater than that of ordinary glass, hence critical angle C for diamond is much smaller (24°) as compared to that of glass (42°).
A skilled diamond cutter thus can take the advantage of such large range of angle of incidence available for total internal reflection 24° to 90°. The diamond can be cut with so many faces, to ensure that light entering the diamond does multiple total internal reflections before coming out. This behavior produce brilliance i.e., sparkling effect in the diamond.

Question 19.
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Let the object be placed x m in front of lens the distance of image from the lens is (3 – x) m.

Question 21.
(a) Determine the ‘effective focal length of the combination of two lenses in question 10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangement. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
(i) Let a parallel beam of light incident first on convex lens, refraction at convex lens

Δθ = Δθ1 + Δθ2
Δθ = A1 (μv1 – μR1) + A2 (μv2 – μR2)

Question 24.
For a normal eye, the far points at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Here the least converging power of eye lens is given as 20 dioptres behind the cornea. If we can calculate the maximum converging power, then we can get the range of accommodation of a normal eye.
(a) To see objects at infinity, the eye uses its least converging power = 40 + 20 = 60 dioptres
∴ Approximate distance between the retina and the cornea eyelens

Question 25.
Does short-sightedness (myopia) or long¬> sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of r accommodation? If not, what might cause these defects of vision?
A person with normal ability of accommodation may be myopic or hypermetropic due to defective eye structure, j- When the eye ball from front to back gets too elongated the myopic defect occur, similarly when the eye ball from front to back gets too shortened the hypermetropia defect occur.
When the eye ball has normal length ‘ but the eye lens loses partially its ability of accommodation, the defect is called “presbyopia” and is corrected in the same I manner as myopia or hypermetropia.

Question 26.
A myopic person has been using spectacles of power – 1.0 dioptre for distant vision. During old age, he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.
Myopic person uses spectacles of power – 1.0 dioptre or concave lens of focal length ƒ =1p  = -100 cm in order to observe clearly object at infinity. Far point of the person can be calculated as

Question 28.
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
(a) At closest distance of the object the image is formed at least distance of distinct vision and eye is most strained.

(c) If we decrease focal length, the lens has to be thick with smaller radius of curvature. In a thick lens both the spherical aberrations and chromatic aberrations become pronounced. Further, grinding for small focal length is not easy. Practically we can not get magnifying power more than 3 with a simple convex lens.

 Magnifying power of a compound microscope is given by

Question 37. ( ncert solution ray optics )
Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in figure. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

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