Light – Reflection and Refraction – Numerical Problems with Stepwise Solutions
Class 10
Science — Physics
Chapter 9: Light — Reflection & Refraction
20 Topic-wise Numerical Problems with Step-by-Step Solutions
Designed strictly as per NCERT Class 10 Chapter 9: Reflection and Refraction — ideal for CBSE board preparation.
How to use: Problems are grouped topic-wise (Reflection, Plane & Spherical Mirrors, Mirror Formula & Magnification, Refraction & Snell's law, Critical angle & TIR, Apparent depth & Practicals). Each problem shows structured steps and final answer. Use Ctrl+P or the "Print / Save as PDF" button below for a printable copy.
Reflection — Plane Mirror (Problems 1–3)
1
A ray of light strikes a plane mirror at an angle of incidence 40°. Find the angle of reflection. (Use reflection law)
Step 1: Law of reflection states: angle of incidence = angle of reflection.
Step 2: Given angle of incidence = 40°.
Answer: Angle of reflection = 40°.
2
An object stands 50 cm in front of a plane mirror. At what distance from the object is its image formed?
Step 1: For a plane mirror, image distance from mirror equals object distance from mirror.
Step 2: Object distance = 50 cm ⇒ image distance from mirror = 50 cm.
Step 3: Distance between object and image = object distance + image distance = 50 + 50 = 100 cm.
Answer: The image is 100 cm away from the object (image is 50 cm behind the mirror).
3
A student stands 1.8 m in front of a plane mirror. How far must they move back so that their image appears 2.6 m away from them?
Step 1: Let object distance = u (from mirror). Image distance = u. Distance between object and image = u + u = 2u.
Step 2: We need 2u = 2.6 m ⇒ u = 1.3 m. Initially u₀ = 1.8 m.
Step 3: Required move back distance = new u − initial u = 1.3 − 1.8 = −0.5 m (negative indicates current position is farther).
Interpretation: actually, since initial distance 1.8 m already gives image distance 3.6 m between object & image; to make object-image separation 2.6 m, the student must move closer to mirror by 0.5 m.
Answer: The student must move the mirror (i.e., 0.5 m closer). Final distance from mirror will be 1.3 m.
Spherical Mirrors — Concave & Convex (Problems 4–7)
4
A concave mirror has a focal length of 15 cm. Where will a parallel beam of light (parallel to principal axis) converge after reflection?
Step 1: Parallel rays to principal axis converge at the principal focus (F).
Step 2: Given focal length f = 15 cm ⇒ focus is at 15 cm from pole along axis.
Answer: They converge at the principal focus, at 15 cm from the mirror.
5
A convex mirror has radius of curvature 30 cm. What is its focal length?
Step 1: For spherical mirrors, focal length f = R/2 (R is radius of curvature). For convex mirror, f = +R/2 in magnitude but image is virtual.
Step 2: R = 30 cm ⇒ f = 30/2 = 15 cm.
Answer: Focal length = 15 cm (virtual focus for convex mirror).
6
An object is placed 10 cm in front of a concave mirror whose focal length is 15 cm. Determine the nature (real/virtual) and position of the image (use the mirror formula).
Step 1: Use mirror formula:
1/f = 1/v + 1/u. Here take object distance u = −10 cm (convention: object in front is negative), focal length f = −15 cm for concave (following sign convention). To avoid sign confusion we use magnitudes and standard algebraic form: 1/f = 1/v + 1/u.Step 2: Use values as magnitudes with understanding that a concave mirror with object between pole and focus gives virtual image. For quick calculation we use formula in magnitudes: 1/|f| = 1/|v| − 1/|u| when object is between pole and focus; equivalently compute using algebraic signs: 1/(−15) = 1/v + 1/(−10) ⇒ −1/15 = 1/v − 1/10.
Step 3: Solve for 1/v: 1/v = −1/15 + 1/10 = (−2 + 3)/30 = 1/30 ⇒ v = 30 cm.
Step 4: Sign: v = +30 cm (positive means image is behind mirror ⇒ virtual image for chosen sign convention). Image is virtual and erect, located 30 cm behind the mirror.
Answer: Virtual, erect image at 30 cm behind the mirror (magnified).
7
An object is placed at the center of curvature (C) of a concave mirror. If radius of curvature is 40 cm, where is the image located and what is its size relative to the object?
Step 1: Center of curvature C is at R = 40 cm ⇒ object distance u = R = 40 cm. Focal length f = R/2 = 20 cm.
Step 2: For object at C (u = R), image is formed at C (v = R), image is real, inverted and of same size.
Answer: Image at 40 cm (at C), real, inverted and of same size as object.
Mirror Formula & Magnification (Problems 8–11)
8
Using mirror formula, find the image distance for an object placed 30 cm from a concave mirror of focal length 10 cm.
Step 1: Mirror formula:
1/f = 1/v + 1/u. Take magnitudes (with sign convention algebra): f = −10 cm (concave), u = −30 cm. 1/(−10) = 1/v + 1/(−30).Step 2: Compute algebraically: −1/10 = 1/v − 1/30 ⇒ 1/v = −1/10 + 1/30 = (−3 + 1)/30 = −2/30 = −1/15 ⇒ v = −15 cm.
Step 3: v = −15 cm (negative → image is real and in front of mirror at 15 cm).
Answer: Image distance = 15 cm in front of the mirror (real, inverted).
9
Find linear magnification for the situation in Problem 8 (u = 30 cm, v = 15 cm). State whether image is magnified or diminished.
Step 1: Linear magnification m = v/u (using magnitudes with sign indicates inversion). Using magnitudes v = 15 cm, u = 30 cm.
Step 2: m = 15/30 = 1/2 = 0.5. Because sign is negative for an inverted image, m = −0.5 (inversion shown by sign).
Answer: Magnification = 0.5 (image is half the size) — image is diminished and inverted.
10
An object 2 cm tall is placed 20 cm from a concave mirror of focal length 10 cm. Find the image height.
Step 1: Given u = 20 cm, f = 10 cm. Mirror formula: 1/f = 1/v + 1/u → 1/10 = 1/v + 1/20 ⇒ 1/v = 1/10 − 1/20 = (2 − 1)/20 = 1/20 ⇒ v = 20 cm.
Step 2: Linear magnification m = v/u = 20/20 = 1. Sign indicates inversion when real, so m = −1.
Step 3: Image height = m × object height = (−1) × 2 cm = −2 cm → magnitude 2 cm (inverted).
Answer: Image height = 2 cm (inverted, same size as object).
11
A convex mirror produces an image that is 1/3 the height of an object placed in front of it. If the object is 60 cm from the mirror, find the image distance.
Step 1: Magnification for mirrors m = v/u (for convex mirror using sign convention v positive behind mirror, u negative). Use magnitudes: |m| = |v|/|u| = 1/3.
Step 2: |v| = |m| × |u| = (1/3) × 60 cm = 20 cm.
Step 3: For convex mirror, image is virtual behind mirror; so image distance = 20 cm behind the mirror.
Answer: Image distance = 20 cm (virtual).
Refraction & Snell's Law (Problems 12–15)
12
A ray of light in air is incident on a glass surface (n = 1.5) at 30°. Find the angle of refraction inside glass. (Use n₁ sin i = n₂ sin r)
Step 1: Given n₁ (air) = 1.00, n₂ (glass) = 1.50, angle of incidence i = 30°.
Step 2: Snell's law: n₁ sin i = n₂ sin r ⇒ 1 × sin30° = 1.5 × sin r.
Step 3: sin30° = 0.5 ⇒ 0.5 = 1.5 sin r ⇒ sin r = 0.5 ÷ 1.5 = 0.333333... ≈ 1/3.
Step 4: r = arcsin(0.333333...) ≈ 19.47° (rounded to two decimals).
Answer: Angle of refraction ≈ 19.47°.
13
A beam of light inside glass (n = 1.5) strikes the glass–air surface at 45°. Will it undergo total internal reflection? (Find critical angle)
Step 1: Critical angle c satisfies sin c = n₂ / n₁ where n₂ is refractive index of rarer medium (air = 1.00) and n₁ is for denser medium (glass = 1.50).
Step 2: sin c = 1.00 / 1.50 = 0.666666... ≈ 2/3.
Step 3: c = arcsin(0.666666...) ≈ 41.81° (rounded to two decimals).
Step 4: Given incident angle 45° which is > critical angle (45° > 41.81°), so total internal reflection occurs.
Answer: Critical angle ≈ 41.81°. Since 45° > 41.81°, the beam undergoes total internal reflection.
14
Light travels from water (n = 4/3) to air. If the angle in water is 48°, what is the angle in air? (Use sin rule)
Step 1: n₁ = 4/3 ≈ 1.333333..., n₂ = 1.00, i = 48° in water.
Step 2: Snell's law: n₁ sin i = n₂ sin r ⇒ (4/3) × sin48° = 1 × sin r.
Step 3: Compute sin48° ≈ 0.743145. Multiply by 4/3: (4/3) × 0.743145 ≈ 0.99086.
Step 4: sin r ≈ 0.99086 ⇒ r ≈ arcsin(0.99086) ≈ 82.56° (approx). This is less than 90°, valid; note if sin r >1 then TIR would occur.
Answer: Angle in air ≈ 82.56°.
15
A ray of light in air strikes a glass slab (n = 1.5) at 60°. Find the angle of refraction inside the slab and emergent ray direction (qualitative).
Step 1: n₁ = 1.00, n₂ = 1.50, i = 60°. Use Snell: sin r = (n₁ / n₂) sin i = (1/1.5) × sin60°.
Step 2: sin60° = √3/2 ≈ 0.866025. Multiply: sin r ≈ 0.866025 ÷ 1.5 = 0.57735.
Step 3: r = arcsin(0.57735) ≈ 35.26° (approx).
Step 4: On emerging back to air from the other side, the emergent ray will be parallel to the original incident ray (for a plane parallel slab), displaced laterally but in the same direction.
Answer: Angle of refraction inside glass ≈ 35.26°. Emergent ray is parallel to the incident ray (laterally displaced).
Critical Angle & Total Internal Reflection (Problems 16–17)
16
Calculate the critical angle for water–air interface. (Use n_water = 4/3)
Step 1: sin c = n₂ / n₁ = n_air / n_water = 1.00 ÷ (4/3) = 3/4 = 0.75.
Step 2: c = arcsin(0.75) ≈ 48.59° (rounded to two decimals).
Answer: Critical angle ≈ 48.59° for the water–air interface.
17
Light inside glass (n = 1.6) strikes glass–air surface at 50°. Will it refract or totally reflect? Compute critical angle first.
Step 1: sin c = n_air / n_glass = 1.00 / 1.60 = 0.625.
Step 2: c = arcsin(0.625) ≈ 38.68°.
Step 3: Given incident angle 50° > 38.68° ⇒ total internal reflection occurs.
Answer: Critical angle ≈ 38.68°. Since 50° > 38.68°, the light undergoes total internal reflection.
Apparent Depth, Slabs & Practicals (Problems 18–20)
18
A coin lies at the bottom of a tank of water 2.4 m deep. What is its apparent depth when viewed from above? (Take n_water = 4/3)
Step 1: Apparent depth d' = real depth × (n_air / n_water). For viewing from air into water, n_air = 1.00.
Step 2: d' = 2.4 m × (1.00 ÷ (4/3)) = 2.4 × (3/4) = 2.4 × 0.75.
Step 3: 2.4 × 0.75 = 1.8 (calculation: 2.4 × 75/100 = (2.4 × 3)/4 = 7.2/4 = 1.8).
Answer: Apparent depth = 1.8 m.
19
A glass slab of thickness 6 cm (n = 1.5) is placed on a book. Find the apparent rise in the position of the book's surface as seen from above.
Step 1: Apparent thickness t' = real thickness × (n_air / n_glass) = t × (1 / 1.5) = t ÷ 1.5.
Step 2: Real thickness t = 6 cm ⇒ apparent thickness t' = 6 ÷ 1.5 = 4 cm.
Step 3: Apparent rise (amount book appears raised) = t − t' = 6 − 4 = 2 cm.
Answer: Apparent rise = 2 cm.
20
A ray of light in air strikes a thin glass slab (n = 1.6) at 45° and inside it the angle is 28°. If the emergent ray is parallel to the incident ray, find the lateral displacement qualitatively and factors it depends on.
Step 1: For a plane parallel slab, emergent ray is parallel to incident ray but shifted laterally. Exact lateral displacement δ depends on slab thickness t, angle of incidence i and refractive index n: δ = t × sin(i − r) / cos r (derived from geometry).
Step 2: Given i = 45°, r = 28° (consistent values). If a numeric thickness t were given, plug in to compute δ. Here we describe dependence:
• δ increases with slab thickness t (directly proportional).
• δ depends on angle difference (i − r) — greater angle of incidence (up to critical) produces larger lateral shift.
• Higher refractive index (larger bending) changes r and so affects δ.
• δ depends on angle difference (i − r) — greater angle of incidence (up to critical) produces larger lateral shift.
• Higher refractive index (larger bending) changes r and so affects δ.
Answer (qualitative): Emergent ray is parallel but laterally displaced. Lateral displacement depends on slab thickness, angle of incidence and refractive index (no numeric value without slab thickness).
Notes & conventions used:
- Mirror formula used:
1/f = 1/v + 1/u. Sign convention: algebraic sign method referenced in worked problems; for conceptual clarity magnitudes are used with explanation of real/virtual location. - Snell's law used as
n₁ sin i = n₂ sin r. Values rounded to two decimal places where necessary. - Refractive indices: air = 1.00, water = 4/3 (≈1.3333). Always state units and direction (behind/in front) for images.
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