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Ray Optics & Optical Instruments NCERT Solutions and Important Questions
In this post, you'll master how light bends, bounces, and forms images through lenses, prisms, and microscopes. This chapter is a high-scoring section for your CBSE 2026 board exams and crucial for competitive tests like JEE and NEET. Tension lene ki zarurat nahi hai (no need to worry)—we have simplified the derivations and numericals for you. Let's dive in!
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Chapter NameRay Optics and Optical Instruments (Chapter 9)
SubjectPhysics
Class12
BoardCBSE (2026-27 Syllabus)
Important TopicsTotal Internal Reflection, Lens Maker's Formula, Prism, Microscopes, Telescopes
Difficulty LevelModerate to Hard (Requires ray diagram practice)
Exam Weightage~9-10 Marks (Combined with Wave Optics for 18 Marks total)
Learning Objectives
After completing this chapter, students will be able to:
- Apply the Cartesian sign convention for spherical mirrors and lenses.
- Understand and derive the relation between focal length and radius of curvature.
- Explain the phenomenon of Total Internal Reflection (TIR) and its applications (like optical fibers).
- Derive and apply the Lens Maker's Formula to solve numerical problems.
- Analyze the refraction of light through a glass prism.
- Calculate the magnifying power of simple microscopes, compound microscopes, and astronomical telescopes.
Key Concepts, Definitions & Formulas
- Laws of Reflection: Angle of incidence equals angle of reflection ($i=r$). The incident ray, reflected ray, and normal lie in the same plane.
- Snell's Law of Refraction: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.
$$n_{21} = \frac{\sin i}{\sin r} = \frac{n_2}{n_1}$$ - Refractive Index ($n$): Speed of light in vacuum divided by speed of light in the medium.
$$n = \frac{c}{v}$$ - Total Internal Reflection (TIR): When light travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle ($i_c$), it reflects completely back into the denser medium.
$$\sin i_c = \frac{n_{rarer}}{n_{denser}}$$ - Lens Maker's Formula: Crucial for manufacturing lenses of specific focal lengths.
$$\frac{1}{f} = (n_{21}-1) \left(\frac{1}{R_1}-\frac{1}{R_2} \right)$$ - Thin Lens Formula:
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ - Power of a Lens ($P$):
$$P = \frac{1}{f (\text{in meters})}$$
(SI Unit: Diopter, D) - Prism Formula:
$$n = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$
(Where $A$ is the angle of prism and $\delta_m$ is the angle of minimum deviation).
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Full NCERT Solutions
Note: Ensure you strictly follow the Cartesian sign convention while solving these.
Question 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?
Answer:
Step 1: Identify Given Values
Height of object, $h_o = +2.5 \text{ cm}$
Object distance, $u = -27 \text{ cm}$
Radius of curvature, $R = -36 \text{ cm}$
Focal length, $f = \frac{R}{2} = -18 \text{ cm}$
Step 2: Find Image Distance ($v$)
Using the mirror formula:
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
$$\frac{1}{-18} = \frac{1}{v} + \frac{1}{-27}$$
$$\frac{1}{v} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27} = \frac{-3+2}{54} = -\frac{1}{54}$$
$$v = -54 \text{ cm}$$
The screen should be placed 54 cm in front of the mirror.
Step 3: Find the Size of the Image ($h_i$)
Magnification, $m = \frac{h_i}{h_o} = -\frac{v}{u}$
$$\frac{h_i}{2.5} = -\frac{-54}{-27} = -2$$
$$h_i = -2 \times 2.5 = -5.0 \text{ cm}$$
Step 4: Describe Nature and Movement
Nature: Real, inverted, and magnified.
If the candle is moved closer (towards $f$), the screen must be moved away from the mirror to catch the sharp image. If it crosses $f$, the image becomes virtual and cannot be caught on a screen.
Step 1: Identify Given Values
Height of object, $h_o = +2.5 \text{ cm}$
Object distance, $u = -27 \text{ cm}$
Radius of curvature, $R = -36 \text{ cm}$
Focal length, $f = \frac{R}{2} = -18 \text{ cm}$
Step 2: Find Image Distance ($v$)
Using the mirror formula:
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
$$\frac{1}{-18} = \frac{1}{v} + \frac{1}{-27}$$
$$\frac{1}{v} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27} = \frac{-3+2}{54} = -\frac{1}{54}$$
$$v = -54 \text{ cm}$$
The screen should be placed 54 cm in front of the mirror.
Step 3: Find the Size of the Image ($h_i$)
Magnification, $m = \frac{h_i}{h_o} = -\frac{v}{u}$
$$\frac{h_i}{2.5} = -\frac{-54}{-27} = -2$$
$$h_i = -2 \times 2.5 = -5.0 \text{ cm}$$
Step 4: Describe Nature and Movement
Nature: Real, inverted, and magnified.
If the candle is moved closer (towards $f$), the screen must be moved away from the mirror to catch the sharp image. If it crosses $f$, the image becomes virtual and cannot be caught on a screen.
Question 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.
Answer:
Step 1: Identify Given Values
$h_o = +4.5 \text{ cm}$
$u = -12 \text{ cm}$
$f = +15 \text{ cm}$ (Convex mirror)
Step 2: Find Image Distance ($v$)
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
$$\frac{1}{15} = \frac{1}{v} + \frac{1}{-12}$$
$$\frac{1}{v} = \frac{1}{15} + \frac{1}{12} = \frac{4+5}{60} = \frac{9}{60}$$
$$v = \frac{60}{9} = +6.67 \text{ cm}$$
Image is formed 6.67 cm behind the mirror (virtual).
Step 3: Find Magnification ($m$)
$$m = -\frac{v}{u} = -\frac{60/9}{-12} = +\frac{5}{9} \approx +0.55$$
Image size $h_i = m \times h_o = \frac{5}{9} \times 4.5 = +2.5 \text{ cm}$.
Step 4: Describe Nature and Movement
Nature: Virtual, erect, and diminished.
As the needle is moved farther, the image size keeps decreasing and shifts towards the focus ($F$).
Step 1: Identify Given Values
$h_o = +4.5 \text{ cm}$
$u = -12 \text{ cm}$
$f = +15 \text{ cm}$ (Convex mirror)
Step 2: Find Image Distance ($v$)
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
$$\frac{1}{15} = \frac{1}{v} + \frac{1}{-12}$$
$$\frac{1}{v} = \frac{1}{15} + \frac{1}{12} = \frac{4+5}{60} = \frac{9}{60}$$
$$v = \frac{60}{9} = +6.67 \text{ cm}$$
Image is formed 6.67 cm behind the mirror (virtual).
Step 3: Find Magnification ($m$)
$$m = -\frac{v}{u} = -\frac{60/9}{-12} = +\frac{5}{9} \approx +0.55$$
Image size $h_i = m \times h_o = \frac{5}{9} \times 4.5 = +2.5 \text{ cm}$.
Step 4: Describe Nature and Movement
Nature: Virtual, erect, and diminished.
As the needle is moved farther, the image size keeps decreasing and shifts towards the focus ($F$).
Question 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?
Answer:
Step 1: Refractive index of water
Given: Real depth $= 12.5 \text{ cm}$, Apparent depth $= 9.4 \text{ cm}$.
$$n_{\text{water}} = \frac{\text{Real Depth}}{\text{Apparent Depth}} = \frac{12.5}{9.4} = 1.33$$
Step 2: New apparent depth with new liquid
Given: $n_{\text{liquid}} = 1.63$, Real depth $= 12.5 \text{ cm}$.
$$\text{New Apparent Depth} = \frac{\text{Real Depth}}{n_{\text{liquid}}} = \frac{12.5}{1.63} \approx 7.67 \text{ cm}$$
Step 3: Distance to move microscope
Distance = Old Apparent Depth - New Apparent Depth
Distance = $9.4 - 7.67 = 1.73 \text{ cm}$.
Final Answer: The microscope must be moved upwards by 1.73 cm.
Step 1: Refractive index of water
Given: Real depth $= 12.5 \text{ cm}$, Apparent depth $= 9.4 \text{ cm}$.
$$n_{\text{water}} = \frac{\text{Real Depth}}{\text{Apparent Depth}} = \frac{12.5}{9.4} = 1.33$$
Step 2: New apparent depth with new liquid
Given: $n_{\text{liquid}} = 1.63$, Real depth $= 12.5 \text{ cm}$.
$$\text{New Apparent Depth} = \frac{\text{Real Depth}}{n_{\text{liquid}}} = \frac{12.5}{1.63} \approx 7.67 \text{ cm}$$
Step 3: Distance to move microscope
Distance = Old Apparent Depth - New Apparent Depth
Distance = $9.4 - 7.67 = 1.73 \text{ cm}$.
Final Answer: The microscope must be moved upwards by 1.73 cm.
Question 9.4: A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Answer:
Step 1: Understand the core concept
Light emerges only through a circular area on the surface. Rays striking at an angle greater than the critical angle ($i_c$) suffer Total Internal Reflection.
Given: $h = 80 \text{ cm} = 0.8 \text{ m}$, $n = 1.33 = \frac{4}{3}$.
Step 2: Calculate Radius ($r$)
Formula for the radius of the circular area:
$$r = \frac{h}{\sqrt{n^2 - 1}}$$
$$r = \frac{0.8}{\sqrt{(1.33)^2 - 1}} = \frac{0.8}{\sqrt{1.77 - 1}} = \frac{0.8}{\sqrt{0.77}} = \frac{0.8}{0.877} \approx 0.912 \text{ m}$$
Step 3: Calculate Area ($A$)
Area of the surface $A = \pi r^2$
$$A = 3.14 \times (0.912)^2 \approx 2.61 \text{ m}^2$$
Final Answer: Light will emerge from a circular area of approximately $2.61 \text{ m}^2$.
Step 1: Understand the core concept
Light emerges only through a circular area on the surface. Rays striking at an angle greater than the critical angle ($i_c$) suffer Total Internal Reflection.
Given: $h = 80 \text{ cm} = 0.8 \text{ m}$, $n = 1.33 = \frac{4}{3}$.
Step 2: Calculate Radius ($r$)
Formula for the radius of the circular area:
$$r = \frac{h}{\sqrt{n^2 - 1}}$$
$$r = \frac{0.8}{\sqrt{(1.33)^2 - 1}} = \frac{0.8}{\sqrt{1.77 - 1}} = \frac{0.8}{\sqrt{0.77}} = \frac{0.8}{0.877} \approx 0.912 \text{ m}$$
Step 3: Calculate Area ($A$)
Area of the surface $A = \pi r^2$
$$A = 3.14 \times (0.912)^2 \approx 2.61 \text{ m}^2$$
Final Answer: Light will emerge from a circular area of approximately $2.61 \text{ m}^2$.
Question 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 $40^\circ$. What is the refractive index of the material of the prism? The refracting angle of the prism is $60^\circ$.
Answer:
Step 1: Identify Given Values
Angle of minimum deviation, $\delta_m = 40^\circ$
Angle of prism, $A = 60^\circ$
Step 2: Apply the Prism Formula
$$n = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$
$$n = \frac{\sin \left(\frac{60^\circ + 40^\circ}{2}\right)}{\sin \left(\frac{60^\circ}{2}\right)}$$
$$n = \frac{\sin(50^\circ)}{\sin(30^\circ)}$$
Given $\sin(50^\circ) \approx 0.766$ and $\sin(30^\circ) = 0.5$
$$n = \frac{0.766}{0.5} = 1.532$$
Final Answer: The refractive index of the prism material is 1.53.
Step 1: Identify Given Values
Angle of minimum deviation, $\delta_m = 40^\circ$
Angle of prism, $A = 60^\circ$
Step 2: Apply the Prism Formula
$$n = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$
$$n = \frac{\sin \left(\frac{60^\circ + 40^\circ}{2}\right)}{\sin \left(\frac{60^\circ}{2}\right)}$$
$$n = \frac{\sin(50^\circ)}{\sin(30^\circ)}$$
Given $\sin(50^\circ) \approx 0.766$ and $\sin(30^\circ) = 0.5$
$$n = \frac{0.766}{0.5} = 1.532$$
Final Answer: The refractive index of the prism material is 1.53.
Extra Important Questions (Board Style 2026)
Multiple Choice Questions (MCQs)
1. When light travels from a denser to a rarer medium, its speed:
a) Decreases
b) Increases
c) Remains constant
d) First increases then decreases
Answer: b) Increases. (Difficulty: Easy)
2. The power of a lens is +2.0 D. Its focal length and nature are:
a) 50 cm, Concave
b) 50 cm, Convex
c) -50 cm, Concave
d) 100 cm, Convex
Answer: b) 50 cm, Convex.
Explanation: $f = \frac{1}{P} = \frac{1}{2} = +0.5 \text{ m} = +50 \text{ cm}$. Positive focal length implies a convex lens. (Difficulty: Easy)
Explanation: $f = \frac{1}{P} = \frac{1}{2} = +0.5 \text{ m} = +50 \text{ cm}$. Positive focal length implies a convex lens. (Difficulty: Easy)
3. Total internal reflection occurs only when the angle of incidence is:
a) Equal to the critical angle
b) Less than the critical angle
c) Greater than the critical angle
d) 90 degrees
Answer: c) Greater than the critical angle. (Difficulty: Medium)
Short Answer Questions (2-3 Marks)
4. Why does a diamond sparkle?
Answer: A diamond sparkles due to Total Internal Reflection. The refractive index of diamond is very high (2.42), making its critical angle very small (about $24.4^\circ$). When light enters a cut diamond, it suffers multiple total internal reflections at various faces before emerging out, causing the brilliant sparkle.
5. Two thin lenses of power +5 D and -3 D are placed in contact. Find the focal length of the combination.
Answer:
Total Power $P = P_1 + P_2 = +5 + (-3) = +2 \text{ D}$.
Focal length $f = \frac{1}{P} = \frac{1}{2} \text{ m} = +50 \text{ cm}$.
Total Power $P = P_1 + P_2 = +5 + (-3) = +2 \text{ D}$.
Focal length $f = \frac{1}{P} = \frac{1}{2} \text{ m} = +50 \text{ cm}$.
6. State two advantages of a reflecting telescope over a refracting telescope.
Answer:
1. Free from chromatic aberration (color blurring) because mirrors are used instead of lenses.
2. Easier to support mechanically since a mirror requires support only at its back, whereas a massive lens needs support at its edges.
1. Free from chromatic aberration (color blurring) because mirrors are used instead of lenses.
2. Easier to support mechanically since a mirror requires support only at its back, whereas a massive lens needs support at its edges.
7. A convex lens is immersed in a liquid having a refractive index equal to that of the lens material. What will happen to its focal length?
Answer: It becomes infinite. The lens will act as a plain glass plate, and the light rays will pass through undeviated.
Long Answer Questions (5 Marks)
8. Derive the Lens Maker's Formula for a convex lens. State the assumptions made.
Answer:
Assumptions: The lens is thin, the aperture is small, and the object is a point on the principal axis.
Step 1: First Surface Refraction
Apply the refraction formula at the first spherical surface:
$$\frac{n_2}{v_1} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1}$$
Step 2: Second Surface Refraction
Apply the refraction formula at the second surface (the image $v_1$ acts as a virtual object):
$$\frac{n_1}{v} - \frac{n_2}{v_1} = \frac{n_1 - n_2}{R_2}$$
Step 3: Combine and Conclude
Add both equations and use the definition of focal length ($v=f$ when $u=\infty$) to arrive at:
$$\frac{1}{f} = (n_{21} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Assumptions: The lens is thin, the aperture is small, and the object is a point on the principal axis.
Step 1: First Surface Refraction
Apply the refraction formula at the first spherical surface:
$$\frac{n_2}{v_1} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1}$$
Step 2: Second Surface Refraction
Apply the refraction formula at the second surface (the image $v_1$ acts as a virtual object):
$$\frac{n_1}{v} - \frac{n_2}{v_1} = \frac{n_1 - n_2}{R_2}$$
Step 3: Combine and Conclude
Add both equations and use the definition of focal length ($v=f$ when $u=\infty$) to arrive at:
$$\frac{1}{f} = (n_{21} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
9. Explain the working of a Compound Microscope with a neat ray diagram. Derive the expression for its magnifying power when the final image is formed at the least distance of distinct vision (D).
Answer:
(Provide a clear ray diagram during exams)
It uses two lenses: an objective (small aperture, small $f$) and an eyepiece (larger aperture).
Magnifying power is given by $m = m_o \times m_e$.
Derivation leads to the formula:
$$m = \left(\frac{L}{f_o}\right) \left(1 + \frac{D}{f_e}\right)$$
Where $L$ is the tube length.
(Provide a clear ray diagram during exams)
It uses two lenses: an objective (small aperture, small $f$) and an eyepiece (larger aperture).
Magnifying power is given by $m = m_o \times m_e$.
Derivation leads to the formula:
$$m = \left(\frac{L}{f_o}\right) \left(1 + \frac{D}{f_e}\right)$$
Where $L$ is the tube length.
10. What is a prism? Derive the relation $n = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$
Answer:
Step 1: Basic setup and angle of deviation
Show angle of deviation $\delta = i + e - A$.
Step 2: Minimum Deviation Conditions
At minimum deviation, $i = e$, and $r_1 = r_2 = r$.
Therefore, $A = 2r \implies r = A/2$.
Also, $\delta_m = 2i - A \implies i = (A + \delta_m)/2$.
Step 3: Apply Snell's Law
Put $i$ and $r$ in Snell's Law ($n = \frac{\sin i}{\sin r}$) to get the final prism formula.
Step 1: Basic setup and angle of deviation
Show angle of deviation $\delta = i + e - A$.
Step 2: Minimum Deviation Conditions
At minimum deviation, $i = e$, and $r_1 = r_2 = r$.
Therefore, $A = 2r \implies r = A/2$.
Also, $\delta_m = 2i - A \implies i = (A + \delta_m)/2$.
Step 3: Apply Snell's Law
Put $i$ and $r$ in Snell's Law ($n = \frac{\sin i}{\sin r}$) to get the final prism formula.
Case-Based Questions
11-13. Read the passage and answer questions:
Optical fibers are extremely thin fibers of glass or plastic used for transmitting optical signals. They work on the principle of Total Internal Reflection. A fiber consists of a core and a cladding. To prevent light from escaping, the refractive index of the materials is carefully chosen.
11. What should be the relation between the refractive index of the core ($n_1$) and the cladding ($n_2$)?
Answer: The refractive index of the core must be greater than that of the cladding ($n_1 > n_2$).
12. Why doesn't the light signal escape the optical fiber even if the fiber is bent?
Answer: Because the light rays strike the core-cladding interface at an angle greater than the critical angle, ensuring continuous Total Internal Reflection.
13. Mention one medical application of optical fibers.
Answer: Endoscopy. Doctors use optical fibers to examine the internal organs of the human body.
Answer: The refractive index of the core must be greater than that of the cladding ($n_1 > n_2$).
12. Why doesn't the light signal escape the optical fiber even if the fiber is bent?
Answer: Because the light rays strike the core-cladding interface at an angle greater than the critical angle, ensuring continuous Total Internal Reflection.
13. Mention one medical application of optical fibers.
Answer: Endoscopy. Doctors use optical fibers to examine the internal organs of the human body.
Assertion-Reason Questions
(Options: a) Both A & R are true, R is correct explanation. b) Both are true, R is not correct explanation. c) A is true, R is false. d) A is false, R is true.)
14. Assertion (A): Air bubble in water shines beautifully.
Reason (R): Due to the refraction of light.
Answer: c) A is true, R is false. (Reason should be Total Internal Reflection, not just refraction).
15. Assertion (A): The focal length of a convex lens decreases when red light is replaced by blue light.
Reason (R): The refractive index of glass is higher for blue light than for red light.
Answer: a) Both A & R are true, and R is the correct explanation. ($\frac{1}{f} \propto (n-1)$, since $n_{\text{blue}} > n_{\text{red}}$, $f_{\text{blue}} < f_{\text{red}}$).
Common Mistakes Students Make
- Sign Convention Errors: This is the #1 reason students lose marks. Always place the object on the left (so $u$ is negative). For convex lenses, $f$ is positive; for concave lenses, $f$ is negative.
- Drawing Sloppy Ray Diagrams: Freehand diagrams without arrows indicating the direction of light will fetch a big ZERO. Use a pencil and scale. Always draw arrows!
- Confusing Formulas: Mirror formula has a "+" ($\frac{1}{v} + \frac{1}{u}$), while the thin lens formula has a "-" ($\frac{1}{v} - \frac{1}{u}$). Write them down before starting the calculation.
- Forgetting to Convert Units: Often, object distance is given in cm but power needs to be calculated. You MUST convert focal length to meters before using $P = \frac{1}{f}$.
Exam Preparation Tips
- Practice Ray Diagrams: Spend 15 minutes daily drawing diagrams for microscopes, telescopes, and prisms. Visual memory works best here.
- Derivations are Gold: The Lens Maker's Formula and Prism Formula are highly expected 3-5 marker questions. Write them out on paper 2-3 times.
- Step-by-Step Numericals: Don't just write the final answer. Write "Given", "Formula Used", "Calculation", and "Final Answer with SI Unit" to secure maximum step marks.
- Formula Sheet: Make a single-page cheat sheet containing all formulas from Chapter 9. Fatafat revise karein right before the exam!
FAQ Section
Is Chapter 9 Ray Optics important for boards?
Absolutely! Combined with Wave Optics, it forms the highest-weightage unit in Class 12 Physics (approx. 18 marks).
Where can I download the Class 12 Physics NCERT PDF?
You can download the latest rationalized 2026-27 NCERT PDFs directly from the official CBSE or NCERT website (ncert.nic.in).
Which questions are most important in Ray Optics?
Lens Maker's Formula derivation, compound microscope ray diagrams, and numericals based on Total Internal Reflection (TIR) and prisms are highly repeated.
What is the difference between a reflecting and refracting telescope?
A refracting telescope uses lenses to gather and focus light, whereas a reflecting telescope uses curved mirrors. Reflecting telescopes are preferred as they avoid color blurring (chromatic aberration).
How to remember sign conventions easily?
Think of the optical center/pole as the origin (0,0) of a graph. Anything measured to the left is negative (-x axis), to the right is positive (+x axis), above the axis is positive (+y), and below is negative (-y).
Conclusion: Ray Optics might seem like a heavy chapter with all its formulas and diagrams, but with logical practice, it becomes the most scoring part of Class 12 Physics. Revise your sign conventions regularly, practice the PYQs (Previous Year Questions), and don't skip the derivations! Download these notes, stay confident, and you'll ace your 2026 board exams. Happy studying!
More Class 12 Physics Chapters
Ch 1: Electric Charges and Fields
Ch 2: Electrostatic Potential and Capacitance
Ch 3: Current Electricity
Ch 4: Moving Charges and Magnetism
Ch 5: Magnetism and Matter
Ch 6: Electromagnetic Induction
Ch 7: Alternating Current
Ch 8: Electromagnetic Waves
Ch 9: Ray Optics
Ch 10: Wave Optics
Ch 11: Dual Nature of Radiation and Matter
Ch 12: Atoms
Ch 13: Nuclei
Class 12 Mock QUIZs
Ch 14: Semiconductor Electronics