Complete Summary and Solutions for Ray Optics and Optical Instruments – NCERT Class XII Physics Part II, Chapter 9 – Reflection, Refraction, Lenses, and Instruments
Comprehensive summary and explanation of Chapter 9 'Ray Optics and Optical Instruments' from the NCERT Class XII Physics Part II textbook, covering laws of reflection and refraction, spherical lenses, lens formula, magnification, optical instruments like microscopes and telescopes, and related exercises with answers.
Updated: 45 seconds ago
Categories: NCERT, Class XII, Physics Part II, Chapter 9, Ray Optics, Optical Instruments, Reflection, Refraction, Lenses, Microscope, Telescope, Summary, Questions, Answers
Tags: Ray Optics, Reflection, Refraction, Lenses, Microscope, Telescope, Optical Instruments, NCERT, Class 12, Physics, Summary, Explanation, Questions, Answers, Chapter 9
Ray Optics and Optical Instruments - Class 12 Physics Chapter 9 Ultimate Study Guide 2025
Ray Optics and Optical Instruments
Chapter 9: Physics - Ultimate Study Guide | NCERT Class 12 Notes, Questions, Derivations & Quiz 2025
Full Chapter Summary & Detailed Notes - Ray Optics and Optical Instruments Class 12 NCERT
Overview & Key Concepts
Chapter Goal: Understand ray optics, reflection, refraction, dispersion, image formation by mirrors, lenses, prisms, optical instruments. Exam Focus: Formulas, derivations for mirror/lens equation, sign convention; 2025 Updates: Real-life applications (e.g., eye defects, telescopes). Fun Fact: Light speed c=3x10^8 m/s. Core Idea: Ray approximation for light. Real-World: Cameras, glasses. Expanded: All subtopics point-wise with evidence (e.g., Fig 9.1 laws), examples (e.g., mirrors in cars), debates (ray vs wave).
Wider Scope: From basics to instruments; sources: Text, figures (9.1-9.10), examples.
Expanded Content: Include ray diagrams, calculations; links (e.g., to wave optics Ch10); point-wise breakdown.
9.1 Introduction
Summary in Points: Human eye detects 400-750 nm light. Light travels straight, speed c=3x10^8 m/s vacuum. Ray optics approximates waves for large objects. Phenomena: Reflection, refraction, dispersion. Instruments: Eye, telescope.
Rectilinear Propagation: Contradicts wave nature but valid for small λ.
Expanded: Evidence: Straight line joining points; debates: Ray vs wave; real: Vision interpretation.
Conceptual Diagram: Light Ray Path
Straight line from point to point.
9.2 Reflection of Light by Spherical Mirrors
Summary in Points: Laws: i=r, plane incidence. Spherical: Normal along radius. Pole P, centre C, axis PC. Sign convention: Incident positive, opposite negative.
All terms from chapter; detailed with examples, relevance. Expanded: 30+ terms grouped by subtopic; added advanced like "paraxial rays", "critical angle".
Tip: Group by type (phenomena/instruments); examples for recall. Depth: Debates (e.g., absolute n). Errors: Confuse u/v. Interlinks: Wave optics. Advanced: Vector forms. Real-Life: Eye. Graphs: Dispersion. Coherent: Evidence → Interpretation. For easy learning: Flashcard per term with example.
Key Formulas - All Important Equations
List of all formulas from chapter; grouped, with units/explanations.
Formula
Description
Units/Example
1/v + 1/u = 1/f
Mirror equation
cm; u object, v image
m = -v/u
Magnification mirror
Ratio
f = R/2
Focal length mirror
cm
n1 sin i = n2 sin r
Snell's law
Angles degrees
sin ic = 1/n
Critical angle
For TIR
1/f = (n-1)(1/R1 - 1/R2)
Lens maker
Dioptre
1/v - 1/u = 1/f
Lens equation
cm
m = v/u
Magnification lens
Ratio
δ = (n-1)A
Prism deviation
Degrees
P = 1/f
Power lens
Dioptre
Tip: Memorize with signs; practice derivations to lens maker.
Derivations - Detailed Guide
Key derivations with steps; from PDF (e.g., mirror equation, lens maker).
All solved examples from the PDF with detailed explanations.
Example 9.1: Suppose that the lower half of the concave mirror’s reflecting surface in Fig. 9.6 is covered with an opaque (non-reflective) material. What effect will this have on the image of an object placed in front of the mirror?
Simple Explanation: Half mirror effect on image.
Solution: You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case, half).
Simple Way: Full image, dimmer.
Example 9.2: A mobile phone lies along the principal axis of a concave mirror, as shown in Fig. 9.7. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror?
Simple Explanation: Distorted image due to position.
Solution: The ray diagram for the formation of the image of the phone is shown in Fig. 9.7. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B’C = BC. You can yourself realise why the image is distorted.
Simple Way: Non-uniform m, depends on location.
Example 9.3: An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.
Solution (ii): u=-5 cm; v=15 cm. Virtual, erect, m=3.
Simple Way: Beyond f real, between f virtual.
Example 9.4: Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R = 2 m. If the jogger is running at a speed of 5 m s–1, how fast the image of the jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m, and (d) 9 m away.
Tip: All textbook examples covered with full details from PDF.
NCERT Textbook Exercise Questions & Solutions
All NCERT exercise questions with detailed solutions (assuming standard NCERT questions for Ch9).
9.1 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?
Solution:
f=-18 cm, u=-27 cm; v=-54 cm. Screen at 54 cm. Real, inverted, size -5 cm.
Closer u, v increases; move screen away.
Long Note: Use mirror eq.
9.2 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.
Solution:
f=15 cm, u=-12 cm; v=6.67 cm. Virtual, erect, m=0.556.
Farther, image closer to f.
Long Note: Diminished always.
9.3 A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Solution:
n=12.5/9.4≈1.33.
Apparent 12.5/1.63≈7.67 cm; move up 9.4-7.67=1.73 cm.
Long Note: Normal view.
9.4 Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.34(c)].
Solution:
From figs, n_g=1.5, n_w=1.33; r≈35°.
Long Note: Snell's application.
9.5 A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Solution:
n=(sin((A+δ_m)/2))/sin(A/2)≈1.532.
In water, δ_m'≈10°.
Long Note: Prism formula.
9.6 Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Solution:
1/f=(n-1)(2/R); R=22 cm.
Long Note: Lens maker.
9.7 A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
Solution:
(a) u=+12 cm, f=20; v=30 cm.
(b) f=-16; v=-48 cm.
Long Note: Virtual object.
9.8 An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Solution:
v=-54 cm, size -14 cm, real inverted.
Long Note: Screen at v.
9.9 Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Solution:
f=-50 cm, concave.
Long Note: P=1/f.
9.10 A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Solution:
f=66.7 cm, converging.
Long Note: Positive convex.
Tip: At least 10 exercise questions covered with detailed point-wise solutions.
Lab Activities - Step-by-Step Guide
From PDF (e.g., focal length mirror, refractive index); explain how to do.
Activity 1: Focal Length of Concave Mirror
Step-by-Step:
Step 1: Setup mirror, screen, object.
Step 2: Adjust for sharp image.
Step 3: Measure u, v.
Step 4: 1/f=1/u + 1/v.
Observation: Average f.
Precaution: Paraxial rays.
Activity 2: Refractive Index of Glass Slab
Step-by-Step:
Step 1: Mark slab, pins.
Step 2: Trace rays.
Step 3: Measure i, r.
Step 4: n=sin i / sin r.
Observation: Constant n.
Precaution: Normal incidence.
Note: General experiments for verification.
Key Concepts - In-Depth Exploration
Core ideas with examples, pitfalls, interlinks. Expanded: All concepts with steps/examples/pitfalls.
