CBSE Class 10 Science Light Reflection and Refraction Numericals 2026

For students preparing for the CBSE Class 10 Science 2026 board exam, the chapter on Light: Reflection and Refraction is one of the highest-scoring yet critical areas in the physics syllabus. Numericals from this chapter regularly appear in 3-mark and 5-mark sections of the question paper. While many students dread calculations involving focal length, magnification, and refractive index, scoring full marks in these questions is straightforward once you master the New Cartesian Sign Convention and understand which formula applies to each optical device. This guide provides an in-depth review of core concepts, complete formula sets, step-by-step solved numericals aligned with the NCERT curriculum, and targeted preparation strategies for your 2026 board examinations.

Key Concepts and Governing Formulas

Before solving any numerical problem, having absolute clarity on the mathematical relationships and sign conventions is paramount. The CBSE marking scheme awards step marks for listing given data, stating the correct formula, substituting values with appropriate signs, and expressing the final answer with its correct unit.

1. The New Cartesian Sign Convention

All distances are measured from the reference origin: the pole (P) in spherical mirrors and the optical centre (O) in spherical lenses.

• Object Distance (u): The object is always placed to the left of the mirror or lens. Therefore, according to convention, u is always negative.
• Focal Length (f):
  • Concave Mirror and Concave Lens: f is always negative.
  • Convex Mirror and Convex Lens: f is always positive.
• Image Distance (v):
  • For Mirrors: Positive if the image is formed behind the mirror (virtual and erect); negative if formed in front of the mirror (real and inverted).
  • For Lenses: Positive if formed on the opposite side of the lens (real and inverted); negative if formed on the same side as the object (virtual and erect).
• Height of Object (h): Measured above the principal axis, so h is always positive.
• Height of Image (h'): Positive for virtual and erect images; negative for real and inverted images.

2. Spherical Mirror Formulas

For a spherical mirror of focal length f and radius of curvature R:

Relation between R and f: R = 2f  or  f = R / 2

Mirror Formula: 1/f = 1/v + 1/u

Linear Magnification (m): m = h' / h = -v / u

A negative magnification signifies a real and inverted image, whereas a positive magnification denotes a virtual and erect image.

3. Refraction and Refractive Index

When light travels obliquely from medium 1 to medium 2, it changes its speed and direction according to Snell's Law:

Snell's Law: sin i / sin r = n₂₁ = constant

Where n₂₁ is the relative refractive index of medium 2 with respect to medium 1:

n₂₁ = (Speed of light in medium 1) / (Speed of light in medium 2) = v₁ / v₂

Absolute Refractive Index (n): n = c / v, where c = 3 × 10⁸ m/s is the speed of light in vacuum and v is the speed of light in the given medium.

4. Spherical Lens Formulas and Power

For a spherical lens of focal length f:

Lens Formula: 1/f = 1/v - 1/u

Linear Magnification (m): m = h' / h = +v / u

Power of a Lens (P): The power of a lens expresses its degree of convergence or divergence of light rays. It is defined as the reciprocal of its focal length expressed in metres:

P = 1 / f (in metres)  or  P = 100 / f (in centimetres)

The SI unit of power is the dioptre (D). A convex lens has positive power, while a concave lens has negative power. When multiple thin lenses are placed in contact, the net power is the algebraic sum: P = P₁ + P₂ + ...

Important CBSE Questions with Answers

The following problems represent standard numerical archetypes frequently set in CBSE Class 10 Science board examinations.

Problem 1: Image Formation by a Concave Mirror

Question: An object 4.0 cm in size is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.

Solution:

1. Identify the given data with sign conventions:
   • Height of object, h = +4.0 cm
   • Object distance, u = -25.0 cm
   • Focal length of concave mirror, f = -15.0 cm
   • Image distance, v = ?
   • Height of image, h' = ?
2. Apply the Mirror Formula:
   1/f = 1/v + 1/u ⇒ 1/v = 1/f - 1/u
   1/v = 1/(-15) - 1/(-25)
   1/v = -1/15 + 1/25
   Taking the LCM of 15 and 25, which is 75:
   1/v = (-5 + 3) / 75 = -2 / 75
   v = -75 / 2 = -37.5 cm
3. Calculate the image size and magnification:
   m = h' / h = -v / u
   h' = - (v × h) / u
   h' = - [(-37.5) × 4.0] / (-25.0)
   h' = - [150.0] / (-25.0) = -6.0 cm

Conclusion: The screen should be placed at a distance of 37.5 cm in front of the mirror. The negative signs of v and h' indicate that the image is real, inverted, and magnified (size 6.0 cm).

Problem 2: Convex Mirror Used as a Rear-View Mirror

Question: A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature, and size of the image.

Solution:

1. Identify the given data:
   • Radius of curvature, R = +3.00 m
   • Focal length, f = R / 2 = +3.00 / 2 = +1.50 m
   • Object distance, u = -5.00 m
2. Use the Mirror Formula:
   1/v = 1/f - 1/u
   1/v = 1/(+1.50) - 1/(-5.00) = 1/1.50 + 1/5.00
   Converting to fractions: 1/v = 2/3 + 1/5 = (10 + 3) / 15 = 13 / 15
   v = 15 / 13 ≈ +1.15 m
3. Calculate Magnification:
   m = -v / u = - (+1.15) / (-5.00) = +1.15 / 5.00 = +0.23

Conclusion: The image is formed at 1.15 m behind the mirror. Because v is positive and m is positive and less than 1, the image is virtual, erect, and diminished to approximately 0.23 times the size of the bus.

Problem 3: Refraction and Absolute Refractive Index

Question: Light enters from air to a glass plate having a refractive index of 1.50. What is the speed of light in the glass? If the refractive index of water is 1.33, calculate the relative refractive index of glass with respect to water.

Solution:

1. Calculate speed of light in glass:
   Given: Refractive index of glass, n_g = 1.50; Speed of light in vacuum, c = 3 × 10⁸ m/s.
   n_g = c / v_g
   v_g = c / n_g = (3 × 10⁸ m/s) / 1.50 = 2.0 × 10⁸ m/s.
2. Calculate relative refractive index:
   Refractive index of glass with respect to water (_wn_g) is given by:
   _wn_g = n_g / n_w = 1.50 / 1.33 ≈ (3/2) / (4/3) = 9/8 = 1.125.

Conclusion: The speed of light in glass is 2.0 × 10⁸ m/s and the refractive index of glass with respect to water is 1.125.

Problem 4: Convex Lens and Real Magnified Image

Question: A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

Solution:

1. Determine object distance and focal length from properties:
   • Image distance, v = +50 cm (real image is formed on the other side of the lens).
   • Since the image is real, inverted, and of the same size as the object, magnification m = -1.
   • Magnification formula for lens: m = v / u ⇒ -1 = 50 / u ⇒ u = -50 cm.
2. Find focal length:
   1/f = 1/v - 1/u = 1/50 - 1/(-50) = 1/50 + 1/50 = 2/50 = 1/25
   f = +25 cm = +0.25 m.
3. Calculate Power of the Lens:
   P = 1 / f (in metres) = 1 / (+0.25) = +4.0 D.

Conclusion: The needle is placed at a distance of 50 cm in front of the lens (at 2F₁), and the power of the lens is +4.0 Dioptres.

Problem 5: Concave Lens and Combination of Thin Lenses

Question: A concave lens has a focal length of 20 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? If this lens is placed in direct contact with a convex lens of focal length 25 cm, calculate the power of the combination.

Solution:

1. Determine object position for the concave lens:
   • Focal length of concave lens, f₁ = -20 cm
   • A concave lens always forms a virtual, erect image on the same side as the object, so v = -10 cm
   Using the Lens Formula:
   1/f = 1/v - 1/u ⇒ 1/u = 1/v - 1/f
   1/u = 1/(-10) - 1/(-20) = -1/10 + 1/20 = (-2 + 1) / 20 = -1 / 20
   u = -20 cm.
2. Calculate the individual powers and combination power:
   Power of concave lens, P₁ = 100 / f₁ (cm) = 100 / (-20) = -5.0 D.
   Power of convex lens, P₂ = 100 / f₂ (cm) = 100 / (+25) = +4.0 D.
   Power of the combination: P = P₁ + P₂ = -5.0 D + 4.0 D = -1.0 D.

Conclusion: The object must be placed at 20 cm in front of the lens. The total power of the lens combination is -1.0 D, meaning the combined system behaves as a diverging lens of focal length -1 m.

How to Prepare for This Topic

Excelling in the numerical section of Chapter 9 requires deliberate practice and methodological consistency. Follow these proven strategies while preparing for your board exam 10 test papers:

1. Avoid Sign Confusion Between Mirrors and Lenses: Remember that the mirror formula has an addition sign (1/f = 1/v + 1/u) with negative magnification (m = -v/u). Conversely, the lens formula has a subtraction sign (1/f = 1/v - 1/u) with positive magnification (m = +v/u). Committing these pairs to memory prevents sign reversal errors.
2. Draw a Mental or Rough Ray Diagram: Whenever you read a question, visualize the position of the object relative to F and 2F (or C). If an object is placed beyond C of a concave mirror, you know in advance that the image must lie between C and F and be diminished. Cross-checking your numerical result against ray diagrams instantly catches calculation mistakes.
3. Convert Focal Length to Metres for Power Calculations: The most frequent error in the board examination is substituting focal length in centimetres directly into P = 1/f. Always convert f to metres first, or use the convenient expression P = 100 / f(in cm).
4. Write Complete Statements with Units: Never stop after arriving at a raw number. Conclude your answer by specifying the position, nature (real/virtual, erect/inverted), size, and appropriate units (cm, m, or D). The CBSE marking scheme dedicates specific marks to the final concluding statement.

Where to Practice More

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