Ray Optics and Optical Instruments Class 12 Physics Numericals 2026-27 | Chapter 9

CBSE Class 12 Physics • Chapter 9 • 2026–27

Ray Optics and Optical Instruments — Numericals with Solutions

Class 12 Physics Chapter 9 Numericals with Solutions for CBSE 2026–27. Practise Ray Optics and Optical Instruments numericals on mirrors, refraction, total internal reflection, optical fibres, spherical surfaces, lenses, lens maker's formula, lens power, prism, microscopes and telescopes.

Study sequence: Revise the Chapter 9 Complete Notes, practise the Chapter 9 Important Questions, attempt the Chapter 9 MCQs, and then use these numericals for calculation practice.

Quick Answer — What do these numericals cover?

This page contains 30 worked numericals plus a separate practice set. The problems move from direct formula application to multi-step application so that you practise the calculation patterns most relevant to the current Chapter 9 scope.

These are original revision problems created for practice. They are not claims about the exact questions that will appear in the CBSE board examination.

1. Current CBSE 2026–27 Numerical Scope

The current CBSE syllabus for Chapter 9 includes spherical mirrors and the mirror formula; refraction and total internal reflection; optical fibres; refraction at spherical surfaces; lenses; thin-lens formula; lens maker's formula; magnification; power; combinations of thin lenses in contact; prism; microscopes; and astronomical telescopes with their magnifying powers. These topics form the basis of the numerical practice below.

Assessment note: Chapter 9 is grouped with Chapter 10 under Unit VI: Optics, which carries 18 marks in the 70-mark Physics theory paper. The CBSE curriculum does not assign a separate fixed mark total to Chapter 9, so this page does not claim a chapter-specific weightage.

Official CBSE Physics Curriculum 2026–27

What should you practise first?

Search behaviour and current Class 12 Ray Optics resources consistently centre on step-by-step numerical solutions, NCERT-style exercise practice, sign-convention problems, formula selection, prism/TIR applications and optical-instrument magnification. This page therefore prioritises those calculation patterns instead of filling the page with repetitive one-step substitutions.

Numerical type	Core skill	Where to practise
Mirror and lens image formation	Sign convention + formula selection	Sections 3 and 5
Snell's law, critical angle and TIR	Refractive-index reasoning	Section 4
Lens maker's formula and power	Unit conversion + algebra	Section 5
Prism minimum deviation	Angle relations + trigonometry	Section 6
Microscope and telescope	Magnifying-power relations	Section 7
Mixed application	Choosing the right model before calculating	Section 8

NCERT's official exemplar resource lists Unit 9: Ray Optics and Optical Instruments, while current solution resources emphasise step-by-step textbook and exemplar-style practice. The questions here are original practice problems, not reproduced textbook questions.

Numerical method that works:

1. Given2. Sign convention3. Formula4. Substitute5. Unit6. Final statement

Do not jump directly to a calculator. In board-style Physics numericals, the correct formula, sign convention and unit conversion are often as important as the final number.

2. Essential Formula Bank

Spherical mirror: 1/f = 1/v + 1/u   ;   f = R/2   ;   m = −v/u

Thin lens: 1/f = 1/v − 1/u   ;   m = v/u

Refractive index: n = c/v   ;   Snell's law: n₁ sin i = n₂ sin r

Critical angle for a medium to air: sin C = 1/n

Spherical refracting surface: n₂/v − n₁/u = (n₂ − n₁)/R

Lens maker's formula in air: 1/f = (n − 1)(1/R₁ − 1/R₂)

Lens maker's formula in a medium: 1/f = (n_lens/n_medium − 1)(1/R₁ − 1/R₂)

Lens power: P = 1/f (f in metre)   ;   lenses in contact: P = P₁ + P₂ + ...

Prism at minimum deviation: n = sin[(A + δ_m)/2] / sin(A/2)

Simple microscope: M_∞ = D/f   ;   M_D = 1 + D/f

Astronomical telescope, normal adjustment: |M| = f_o/f_e   ;   L = f_o + f_e

Use the Cartesian sign convention consistently. Take D = 25 cm for the least distance of distinct vision when the question does not specify another value.

How to write a board-style numerical answer

1. Given: Write the known quantities with signs and units.
2. Sign convention: State the signs of u, v, f, R or other relevant quantities before substitution.
3. Formula: Write the governing equation before inserting numbers.
4. Calculation: Show the key algebraic step instead of giving only the final value.
5. Interpretation: State whether the image is real/virtual, erect/inverted, enlarged/diminished, or whether TIR occurs.
6. Unit: Finish with the correct physical unit and sensible rounding.

This format also makes it easier to diagnose whether an error came from physics, signs, algebra or units.

3. Reflection and Spherical Mirrors

1. A concave mirror has a radius of curvature of 40 cm. An object is placed 60 cm in front of it. Find the image position and magnification.

Given: R = −40 cm, u = −60 cm. Therefore f = R/2 = −20 cm.

Solution:

1/f = 1/v + 1/u

1/v = 1/f − 1/u = −1/20 − (−1/60) = −1/30

Therefore v = −30 cm.

m = −v/u = −(−30)/(−60) = −0.5.

Answer: Image forms 30 cm in front of the mirror; it is real and inverted, with magnification −0.5.

2. A convex mirror has focal length 15 cm. An object is placed 30 cm in front of it. Find the image distance and magnification.

Given: f = +15 cm, u = −30 cm.

Solution:

1/v = 1/f − 1/u = 1/15 + 1/30 = 1/10

Therefore v = +10 cm.

m = −v/u = −10/(−30) = +1/3.

Answer: Image is 10 cm behind the mirror and is virtual, erect and diminished.

3. A concave mirror of focal length 20 cm forms a real image 60 cm in front of the mirror. Find the object distance and magnification.

Given: f = −20 cm, v = −60 cm.

Solution:

1/u = 1/f − 1/v = −1/20 − (−1/60) = −1/30

Therefore u = −30 cm.

m = −v/u = −(−60)/(−30) = −2.

Answer: Object is 30 cm in front of the mirror; the image is inverted and twice the object size.

4. A 4 cm tall object is placed 60 cm in front of the concave mirror of Question 1. Find the image height.

Given: Object height h_o = 4 cm and m = −0.5.

Solution:

m = h_i/h_o

h_i = m h_o = −0.5 × 4 = −2 cm.

Answer: Image height = −2 cm; the negative sign indicates inversion.

5. A concave mirror forms an image at the same distance from the mirror as the object. If the object is 30 cm in front of the mirror, find the focal length.

Given: u = −30 cm and v = −30 cm.

Solution:

1/f = 1/v + 1/u = −1/30 − 1/30 = −1/15

Therefore f = −15 cm.

Answer: The focal length is −15 cm. The object is at the centre of curvature.

4. Refraction, Refractive Index and Total Internal Reflection

6. Light travels through a transparent medium at 2 × 10⁸ m/s. Find its refractive index.

Given: c = 3 × 10⁸ m/s and v = 2 × 10⁸ m/s.

Solution:

n = c/v

n = (3 × 10⁸)/(2 × 10⁸) = 1.5.

Answer: Refractive index = 1.5.

7. A ray travels from glass of refractive index 1.5 into air at an angle of incidence 30°. Find the angle of refraction.

Given: n₁ = 1.5, n₂ = 1, i = 30°.

Solution:

n₁ sin i = n₂ sin r

sin r = 1.5 × sin 30° = 1.5 × 0.5 = 0.75

Therefore r = sin⁻¹(0.75) ≈ 48.6°.

Answer: Angle of refraction ≈ 48.6°.

8. Find the critical angle for a glass-air interface when the refractive index of glass is 1.5.

Given: n = 1.5.

Solution:

sin C = 1/n = 1/1.5 = 0.6667

Therefore C ≈ 41.8°.

Answer: Critical angle ≈ 41.8°.

9. A ray inside glass of refractive index 1.5 strikes a glass-air boundary at 50°. Will total internal reflection occur?

Given: Critical angle for glass-air ≈ 41.8°; incidence angle = 50°.

Solution:

For TIR, light must travel from denser to rarer medium and i > C.

Here glass → air and 50° > 41.8°.

Answer: Yes. Total internal reflection occurs.

10. A transparent medium has refractive index 1.33 with respect to air. Find its critical angle approximately.

Given: n = 1.33.

Solution:

sin C = 1/1.33 ≈ 0.752

Therefore C ≈ 48.8°.

Answer: Critical angle ≈ 48.8°.

11. A liquid has real depth 12 cm and apparent depth 9 cm when viewed normally from air. Find its refractive index.

Given: Real depth = 12 cm; apparent depth = 9 cm.

Solution:

n = real depth / apparent depth

n = 12/9 = 1.33.

Answer: Refractive index ≈ 1.33.

5. Refraction at Spherical Surfaces and Lenses

12. A convex lens of focal length 20 cm has an object placed 30 cm in front of it. Find the image distance and magnification.

Given: f = +20 cm, u = −30 cm.

Solution:

1/f = 1/v − 1/u

1/v = 1/f + 1/u = 1/20 − 1/30 = 1/60

Therefore v = +60 cm.

m = v/u = 60/(−30) = −2.

Answer: Image forms 60 cm on the other side of the lens and is real, inverted and twice the object size.

13. A concave lens has focal length −15 cm. An object is placed 30 cm in front of it. Find the image distance and magnification.

Given: f = −15 cm, u = −30 cm.

Solution:

1/v = 1/f + 1/u = −1/15 − 1/30 = −1/10

Therefore v = −10 cm.

m = v/u = (−10)/(−30) = +1/3.

Answer: Image forms 10 cm in front of the lens on the object side and is virtual, erect and diminished.

14. A convex lens forms a real image 30 cm from the lens when the object is 60 cm in front of it. Find its focal length.

Given: u = −60 cm, v = +30 cm.

Solution:

1/f = 1/v − 1/u = 1/30 + 1/60 = 1/20

Therefore f = +20 cm.

Answer: Focal length = +20 cm.

15. A plano-convex glass lens has refractive index 1.5 and the radius of curvature of its curved surface is 20 cm. Find its focal length in air.

Given: n = 1.5, R₁ = +20 cm, R₂ = ∞.

Solution:

1/f = (n − 1)(1/R₁ − 1/R₂)

1/f = 0.5(1/20 − 0) = 1/40

Therefore f = +40 cm.

Answer: Focal length = +40 cm.

16. A convex lens has power +5 D. Find its focal length.

Given: P = +5 D.

Solution:

P = 1/f, where f is in metre.

f = 1/5 = 0.20 m = 20 cm.

Answer: Focal length = +20 cm.

17. A concave lens has focal length −50 cm. Find its power.

Given: f = −0.50 m.

Solution:

P = 1/f = 1/(−0.50) = −2 D.

Answer: Power = −2 D.

18. Two thin lenses of powers +4 D and −1.5 D are placed in contact. Find the equivalent power and focal length.

Given: P₁ = +4 D, P₂ = −1.5 D.

Solution:

P = P₁ + P₂ = 4 − 1.5 = +2.5 D.

f = 1/P = 1/2.5 = 0.40 m = 40 cm.

Answer: Equivalent power = +2.5 D; focal length = +40 cm.

19. A convex lens has refractive index 1.5 and surface radii R₁ = +20 cm and R₂ = −20 cm. Find its focal length in air.

Given: n = 1.5, R₁ = +20 cm, R₂ = −20 cm.

Solution:

1/f = (1.5 − 1)[1/20 − (−1/20)]

1/f = 0.5 × 2/20 = 1/20

Therefore f = +20 cm.

Answer: Focal length = +20 cm.

20. A glass spherical surface separates air (n₁ = 1) from glass (n₂ = 1.5). Its radius of curvature is +20 cm. An object is 30 cm in front of the surface. Find the image distance.

Given: n₁ = 1, n₂ = 1.5, u = −30 cm, R = +20 cm.

Solution:

Use n₂/v − n₁/u = (n₂ − n₁)/R.

1.5/v − 1/(−30) = 0.5/20

1.5/v + 1/30 = 1/40

1.5/v = −1/120

Therefore v = −180 cm.

Answer: The image is virtual and lies 180 cm on the object side of the spherical surface.

21. A glass lens of refractive index 1.5 is immersed in a liquid of refractive index 1.33. For a plano-convex lens with curved radius 20 cm, estimate its focal length in the liquid.

Given: n_lens = 1.5, n_medium = 1.33, R₁ = +20 cm, R₂ = ∞.

Solution:

1/f = (1.5/1.33 − 1)(1/20)

The factor 1.5/1.33 − 1 ≈ 0.1278.

Therefore 1/f ≈ 0.00639 cm⁻¹.

Hence f ≈ +156.5 cm.

Answer: Focal length in the liquid is approximately +1.57 m. The lens becomes much weaker than in air.

6. Prism Numericals

22. A prism has angle A = 60° and minimum deviation δ_m = 40°. Find its refractive index.

Given: A = 60°, δ_m = 40°.

Solution:

n = sin[(A + δ_m)/2] / sin(A/2)

n = sin 50° / sin 30°

n ≈ 0.7660/0.5 = 1.53.

Answer: Refractive index ≈ 1.53.

23. A prism has refractive index 1.5 and prism angle 60°. Find the minimum deviation.

Given: n = 1.5, A = 60°.

Solution:

n = sin[(A + δ_m)/2] / sin(A/2)

1.5 = sin[(60° + δ_m)/2]/sin30°

Therefore sin[(60° + δ_m)/2] = 0.75.

(60° + δ_m)/2 ≈ 48.59°.

Thus δ_m ≈ 37.18°.

Answer: Minimum deviation ≈ 37.2°.

24. At minimum deviation, a prism has A = 60°. If the refractive index is 1.5, find the angle of refraction at each face.

Given: At minimum deviation, r₁ = r₂ and r₁ + r₂ = A.

Solution:

2r = A = 60°

Therefore r = 30° at each face.

Answer: Each internal refraction angle is 30°.

25. For an equilateral prism at minimum deviation, the angle of incidence is 50° and the angle of emergence is also 50°. Find the minimum deviation.

Given: i = 50°, e = 50°, A = 60°.

Solution:

δ = i + e − A

δ = 50° + 50° − 60° = 40°.

Answer: Minimum deviation = 40°.

7. Optical Instruments — Microscopes and Telescopes

26. A simple microscope has focal length 5 cm. Find its magnifying power when the final image is formed at infinity. Take D = 25 cm.

Given: f = 5 cm, D = 25 cm.

Solution:

M_∞ = D/f = 25/5 = 5.

Answer: Magnifying power = 5.

27. A simple microscope has focal length 5 cm. Find its magnifying power when the final image is formed at the least distance of distinct vision.

Given: f = 5 cm, D = 25 cm.

Solution:

M_D = 1 + D/f

M_D = 1 + 25/5 = 6.

Answer: Magnifying power = 6.

28. An astronomical telescope is adjusted for normal vision. Its objective focal length is 90 cm and eyepiece focal length is 10 cm. Find its magnifying power and tube length.

Given: f_o = 90 cm, f_e = 10 cm.

Solution:

|M| = f_o/f_e = 90/10 = 9. The negative sign is used when the angular magnification is written with image orientation.

L = f_o + f_e = 90 + 10 = 100 cm.

Answer: Magnifying power = 9; tube length = 100 cm.

29. An astronomical telescope in normal adjustment has magnifying power 12 and tube length 130 cm. Find the focal lengths of the objective and eyepiece.

Given: M = 12 and L = 130 cm.

Solution:

|M| = f_o/f_e = 12, so f_o = 12f_e.

L = f_o + f_e = 130 cm.

12f_e + f_e = 130 ⇒ 13f_e = 130.

Therefore f_e = 10 cm and f_o = 120 cm.

Answer: Objective focal length = 120 cm; eyepiece focal length = 10 cm.

30. A compound microscope has objective focal length 1 cm, eyepiece focal length 5 cm and tube length 20 cm. Estimate its magnifying power for normal adjustment using M = (L/f_o)(D/f_e), with D = 25 cm.

Given: f_o = 1 cm, f_e = 5 cm, L = 20 cm, D = 25 cm.

Solution:

M = (20/1)(25/5)

M = 20 × 5 = 100.

Answer: Approximate magnifying power = 100.

8. Mixed Board-Style Challenge Numericals

Challenge 1. A concave mirror of focal length 15 cm is used to form a real image twice the size of the object. Find the object and image distances.

Given: f = −15 cm and for a real inverted image m = −2.

Solution:

m = −v/u = −2 ⇒ v = 2u.

Using 1/f = 1/v + 1/u:

−1/15 = 1/(2u) + 1/u = 3/(2u).

Therefore u = −22.5 cm and v = −45 cm.

Answer: Object distance = 22.5 cm in front; image distance = 45 cm in front.

Challenge 2. A convex lens has focal length 15 cm. An object is moved from 30 cm to 20 cm in front of it. Find the image distance in the second position.

Given: f = +15 cm, u = −20 cm.

Solution:

1/v = 1/f + 1/u = 1/15 − 1/20 = 1/60.

Therefore v = +60 cm.

Answer: The image forms 60 cm from the lens on the opposite side.

Challenge 3. A +5 D convex lens is placed in contact with a −7 D concave lens. Find the equivalent focal length and state whether the combination is converging or diverging.

Given: P₁ = +5 D, P₂ = −7 D.

Solution:

P = 5 − 7 = −2 D.

f = 1/P = 1/(−2) = −0.50 m.

Answer: Equivalent focal length = −50 cm; the combination is diverging.

Challenge 4. A ray passes from glass (n = 1.5) to air. Find the incidence angle for which the refracted ray just grazes the surface.

Given: At the limiting condition, r = 90°.

Solution:

n sin C = 1 × sin90°.

sin C = 1/1.5.

Therefore C ≈ 41.8°.

Answer: Incidence angle ≈ 41.8°. This is the critical angle.

Challenge 5. A plano-convex lens has n = 1.5 and curved-surface radius 30 cm. Find its focal length in air.

Given: R₁ = +30 cm, R₂ = ∞.

Solution:

1/f = (1.5 − 1)(1/30) = 1/60.

Therefore f = +60 cm.

Answer: Focal length = +60 cm.

Challenge 6. A telescope has objective focal length 1.2 m and eyepiece focal length 0.10 m. Find the magnifying power and tube length for normal adjustment.

Given: f_o = 1.2 m, f_e = 0.10 m.

Solution:

|M| = f_o/f_e = 1.2/0.10 = 12.

L = f_o + f_e = 1.2 + 0.10 = 1.30 m.

Answer: Magnifying power = 12; tube length = 1.30 m.

9. Additional High-Value Application Numericals

31. A glass optical fibre has core refractive index 1.50 and cladding refractive index 1.45. Find the critical angle at the core-cladding boundary and determine whether a ray incident at 78° to the normal undergoes total internal reflection.

Given: n_core = 1.50, n_cladding = 1.45, i = 78°.

Solution:

For the core-to-cladding boundary, sin C = n_cladding/n_core = 1.45/1.50 = 0.9667.

Therefore C ≈ 75.1°.

Since 78° > 75.1°, the incidence angle is greater than the critical angle.

Answer: Critical angle ≈ 75.1°; total internal reflection occurs.

32. A convex lens of focal length 20 cm forms a real image 40 cm from the lens. Find the object distance and magnification.

Given: f = +20 cm, v = +40 cm.

Solution:

1/f = 1/v − 1/u.

1/u = 1/v − 1/f = 1/40 − 1/20 = −1/40.

Therefore u = −40 cm.

m = v/u = 40/(−40) = −1.

Answer: Object distance = 40 cm in front of the lens; magnification = −1. The image is real, inverted and equal in size.

33. A concave mirror of focal length 20 cm is placed so that a 5 cm object forms a 10 cm real image. Find the object distance and image distance.

Given: f = −20 cm and m = h_i/h_o = −10/5 = −2.

Solution:

m = −v/u = −2 ⇒ v = 2u.

−1/20 = 1/(2u) + 1/u = 3/(2u).

Therefore u = −30 cm and v = −60 cm.

Answer: Object distance = 30 cm in front; image distance = 60 cm in front.

34. A prism has angle A = 60° and refractive index 1.60. Find the minimum deviation.

Given: A = 60°, n = 1.60.

Solution:

n = sin[(A + δ_m)/2]/sin(A/2).

sin[(60° + δ_m)/2] = 1.60 × sin30° = 0.80.

(60° + δ_m)/2 = sin⁻¹(0.80) ≈ 53.13°.

Therefore δ_m ≈ 46.26°.

Answer: Minimum deviation ≈ 46.3°.

35. A simple microscope has focal length 4 cm. Compare its magnifying power for final image at infinity and at the least distance of distinct vision, taking D = 25 cm.

Given: f = 4 cm, D = 25 cm.

Solution:

For final image at infinity: M_∞ = D/f = 25/4 = 6.25.

For final image at D: M_D = 1 + D/f = 1 + 25/4 = 7.25.

Answer: M_∞ = 6.25 and M_D = 7.25. The near-point adjustment gives the larger magnifying power.

10. Practice Set — Try These Yourself

Attempt these without looking at the answers first. Write Given → Sign convention → Formula → Substitution → Answer with unit for every problem.

Practice 1. A concave mirror has focal length 25 cm. An object is placed 50 cm in front of it. Find the image distance and magnification.

Practice 2. A convex mirror has focal length 20 cm and an object is placed 40 cm in front of it. Find the image distance.

Practice 3. A convex lens of focal length 10 cm has an object at 15 cm. Find the image distance and magnification.

Practice 4. A medium has refractive index 1.6. Find its critical angle with respect to air.

Practice 5. A liquid has real depth 15 cm and refractive index 1.5. Find its apparent depth when viewed normally from air.

Practice 6. Two lenses of powers +3 D and +2 D are placed in contact. Find their equivalent focal length.

Practice 7. A biconvex lens has n = 1.5 and radii +20 cm and −20 cm. Find its focal length.

Practice 8. A prism has A = 60° and minimum deviation 30°. Find its refractive index.

Practice 9. A simple microscope has focal length 4 cm. Find its magnifying power for normal adjustment. Take D = 25 cm.

Practice 10. An astronomical telescope has magnifying power 8 and tube length 90 cm in normal adjustment. Find the focal lengths of the objective and eyepiece.

11. Practice Set — Answer Check

Practice	Final answer
1	v = −50 cm; m = −1
2	v = +40/3 cm ≈ +13.3 cm
3	v = +30 cm; m = −2
4	C ≈ 38.7°
5	Apparent depth = 10 cm
6	P = +5 D; f = +20 cm
7	f = +20 cm
8	n = sin45°/sin30° ≈ 1.414
9	M∞ = 25/4 = 6.25
10	fe = 10 cm; fo = 80 cm

12. Common Numerical Mistakes to Avoid

• Using the mirror formula for a lens or the lens formula for a mirror.
• Forgetting that mirror magnification is m = −v/u while thin-lens magnification is m = v/u.
• Using focal length in centimetres directly in P = 1/f without converting f to metres.
• Ignoring the sign of u, v, f and R under the Cartesian convention.
• Applying total internal reflection when light is travelling from rarer to denser medium.
• Forgetting that TIR requires the incidence angle to be greater than the critical angle, not merely equal to it.
• Using the air-form lens maker's formula when the lens is immersed in another medium.
• Forgetting that at minimum deviation in a prism, i = e and r₁ = r₂ = A/2.
• Confusing magnification of a lens/mirror with magnifying power of an optical instrument.
• Reporting a numerical answer without its unit or without interpreting the sign.

13. Quick Answers — Ray Optics Numericals

How do you solve Class 12 Physics Chapter 9 numericals?
Start with the optical system, write the sign convention, select the correct formula, substitute with consistent units, and interpret the sign of the result.

How do you solve a mirror numerical?
Apply the Cartesian sign convention, use 1/f = 1/v + 1/u, solve for v, and then use m = −v/u if magnification is required.

How do you solve a thin-lens numerical?
Use 1/f = 1/v − 1/u with the Cartesian convention, then calculate magnification from m = v/u.

How do you calculate lens power?
Use P = 1/f with focal length in metres. A converging lens has positive power and a diverging lens has negative power.

When does total internal reflection occur?
Light must travel from an optically denser medium to a rarer medium and the incidence angle must be greater than the critical angle.

What happens at minimum deviation in a prism?
The path is symmetric: i = e and r₁ = r₂ = A/2.

What is the magnifying power of a simple microscope for final image at infinity?
M = D/f.

What is the magnifying power of an astronomical telescope in normal adjustment?
The magnitude is |M| = f_o/f_e, while the signed value is commonly written M = −f_o/f_e for the inverted final image. The tube length is L = f_o + f_e.

14. Official CBSE and NCERT Resources

CBSE Physics Curriculum 2026–27

CBSE Class XII 2026–27 Sample Question Papers & Marking Schemes

NCERT Physics Part-II Textbook Portal

NCERT Exemplar Problems

CBSE Previous Years' Question Papers

15. Final Numerical Revision Checklist

• ☐ I can apply the Cartesian sign convention without guessing.
• ☐ I can solve mirror and lens equations step by step.
• ☐ I can calculate magnification and interpret its sign.
• ☐ I can use Snell's law and calculate refractive index.
• ☐ I can calculate critical angle and identify TIR conditions.
• ☐ I can use the spherical refracting-surface relation.
• ☐ I can use lens maker's formula and calculate lens power.
• ☐ I can combine powers of thin lenses in contact.
• ☐ I can solve prism minimum-deviation numericals.
• ☐ I can calculate microscope and telescope magnifying power.
• ☐ I write the final numerical answer with the correct sign and unit.

Continue your Chapter 9 study path: Complete Notes  →  Important Questions  →  MCQs  →  Numericals