This MCQ module is based on: Refraction of Light, Lenses and Power of a Lens
TOPIC 34 OF 50
Refraction of Light, Lenses and Power of a Lens
🎓 Class 10
Science
CBSE
Theory
Ch 9 — Light – Reflection and Refraction
⏱ ~25 min
🌐 Language: [gtranslate]
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9.9 Refraction of Light — Bending at a Boundary
Dip a pencil partly into a glass of water and the submerged part looks bent. A coin placed at the bottom of an opaque mug becomes visible when water is poured in. These everyday surprises happen because light changes direction when it passes from one transparent medium into another. This change in direction is called refraction.
Why refraction happens: Light travels at different speeds in different media. When a ray enters a denser medium (like water from air), it slows down. The change in speed at the boundary bends the ray — just as a car entering soft sand from a hard road slows and turns.
9.9.1 Laws of Refraction — Snell's Law
Law 1: The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
Law 2 (Snell's law): For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant: \[ \dfrac{\sin i}{\sin r} = n_{21} \quad \text{or}\quad n_1\sin i = n_2 \sin r \] This constant is called the refractive index of medium 2 with respect to medium 1.
Law 2 (Snell's law): For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant: \[ \dfrac{\sin i}{\sin r} = n_{21} \quad \text{or}\quad n_1\sin i = n_2 \sin r \] This constant is called the refractive index of medium 2 with respect to medium 1.
9.9.2 Refractive Index
The refractive index of a medium is a number that tells us how much it slows light down.
Absolute refractive index (of a medium, relative to vacuum):
\[ n = \dfrac{\text{speed of light in vacuum}}{\text{speed of light in medium}} = \dfrac{c}{v} \]Relative refractive index of medium 2 with respect to medium 1:
\[ n_{21} = \dfrac{n_2}{n_1} = \dfrac{v_1}{v_2} \]A medium with higher refractive index is said to be optically denser. (Optical density is not the same as mass density — kerosene is lighter than water but optically denser.)
| Medium | Refractive index (n) | Medium | Refractive index (n) |
|---|---|---|---|
| Vacuum | 1.00 (exact) | Crown glass | 1.52 |
| Air (STP) | 1.0003 ≈ 1 | Flint glass | 1.65 |
| Ice | 1.31 | Ruby | 1.71 |
| Water | 1.33 | Sapphire | 1.77 |
| Ethyl alcohol | 1.36 | Diamond | 2.42 |
| Kerosene | 1.44 | Rock salt | 1.54 |
9.10 Refraction Through a Rectangular Glass Slab
When a ray of light passes obliquely through a rectangular glass slab, it bends toward the normal on entering glass (air→glass) and bends away on leaving (glass→air). Because the two surfaces are parallel, the emergent ray is parallel to the incident ray — but shifted sideways. This sideways shift is called the lateral displacement (or lateral shift).
9.11 Refraction by Spherical Lenses
A lens is a piece of transparent material bounded by two surfaces, at least one of which is curved (spherical). Two common types:
- Convex lens (converging): thicker in the middle, thinner at the edges. Parallel rays are converged to a real focus on the other side.
- Concave lens (diverging): thinner in the middle, thicker at the edges. Parallel rays are diverged — they appear to come from a virtual focus on the same side as the object.
9.11.1 Key Terms of a Lens
- Optical centre (O): the centre of the lens. A ray through O goes straight, undeviated.
- Principal axis: the straight line joining the centres of curvature of the two surfaces.
- Principal focus (F): the point on the principal axis where rays parallel to it converge (convex) or appear to diverge from (concave). A lens has two foci, F1 and F2, one on each side, equidistant from O.
- Focal length (f): distance OF.
- Centres of curvature 2F1, 2F2 at twice the focal length on either side.
9.11.2 Standard Rays for Ray Diagrams (Convex Lens)
- A ray parallel to the principal axis passes through F2 after refraction.
- A ray through F1 (before the lens) emerges parallel to the principal axis.
- A ray through the optical centre O goes straight without any bending.
9.11.3 Image Formation by a Convex Lens — Six Positions
Case 1 — Object at infinity
Image at F2; highly diminished (point-sized), real and inverted.
Case 2 — Object beyond 2F1
Image between F2 and 2F2; real, inverted, diminished. (This is how a camera forms its image on the film/sensor.)
Case 3 — Object at 2F1
Image at 2F2; real, inverted, same size as the object.
Case 4 — Object between F1 and 2F1
Image beyond 2F2; real, inverted, enlarged. (Used in slide and movie projectors.)
Case 5 — Object at F1
Refracted rays are parallel — image at infinity, highly enlarged, real and inverted.
Case 6 — Object between O and F1
Image on the same side as the object; virtual, erect, enlarged. (This is how a simple magnifying glass works.)
| Object position | Image position | Size | Nature |
|---|---|---|---|
| At infinity | At F2 | Highly diminished (point) | Real, inverted |
| Beyond 2F1 | Between F2 and 2F2 | Diminished | Real, inverted |
| At 2F1 | At 2F2 | Same size | Real, inverted |
| Between F1 and 2F1 | Beyond 2F2 | Enlarged | Real, inverted |
| At F1 | At infinity | Highly enlarged | Real, inverted |
| Between O and F1 | Same side as object | Enlarged | Virtual, erect |
9.11.4 Image Formation by a Concave Lens
A concave lens always gives a virtual, erect, diminished image located between the optical centre and the principal focus F1, on the same side as the object — no matter where the object is placed.
9.12 Lens Formula and Magnification
Lens formula (with sign convention: distances measured from O; object distance \(u\) negative; for a convex lens f is positive, for a concave lens f is negative):
\[ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \]
Magnification by a lens:
\[ m = \dfrac{h'}{h} = \dfrac{v}{u} \]
Note the plus sign, unlike the mirror case.
9.13 Power of a Lens
A lens with a short focal length bends light more strongly. This "bending ability" is called the power of the lens, denoted by \(P\):
\[ P = \dfrac{1}{f\,(\text{in metres})} \]The SI unit of power is the dioptre (D): \(1\ \text{D} = 1\ \text{m}^{-1}\). A convex lens has positive power; a concave lens has negative power.
When two lenses of powers \(P_1\) and \(P_2\) are placed in contact, the combined power is \(P = P_1 + P_2\).
9.14 Worked Numerical Examples
Example 1 — Speed and refractive index
Given: Refractive index of water is 1.33. Find the speed of light in water. (c = 3 × 10⁸ m/s)
\[ n = c/v \Rightarrow v = c/n = \dfrac{3\times 10^{8}}{1.33} \approx 2.26\times 10^{8}\ \text{m/s} \]
Example 2 — Snell's law
Given: A ray enters glass (n = 1.5) from air at an angle of incidence 45°. Find the angle of refraction.
\[ n_{\text{glass}} = \dfrac{\sin i}{\sin r} \Rightarrow \sin r = \dfrac{\sin 45°}{1.5} = \dfrac{0.707}{1.5} = 0.471 \]
\(r = \sin^{-1}(0.471) \approx 28.1°\). The ray bends toward the normal.
Example 3 — Convex lens, beyond 2F₁
Given: A 4 cm tall object is placed 30 cm in front of a convex lens of focal length 10 cm. Find image position, nature and size.
\(u = -30\) cm, \(f = +10\) cm.
\[ \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{10} + \dfrac{1}{-30} = \dfrac{3-1}{30} = \dfrac{2}{30} = \dfrac{1}{15} \]\(v = +15\) cm (real image on opposite side).
\(m = v/u = 15/(-30) = -0.5\). Height \(h' = mh = -0.5 \times 4 = -2\) cm.
Real, inverted, diminished, 2 cm tall, 15 cm from the lens on the other side.
Example 4 — Convex lens used as a magnifier
Given: A convex lens has focal length 15 cm. An object 3 cm tall is placed 10 cm in front of it. Find the image.
\(u=-10\) cm, \(f=+15\) cm.
\[ \dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{-10} = \dfrac{2-3}{30} = -\dfrac{1}{30} \]\(v = -30\) cm (virtual, same side as object).
\(m = v/u = (-30)/(-10) = +3\). Image is 9 cm tall, erect, virtual.
A simple magnifying glass at work.
Example 5 — Concave lens
Given: A concave lens of focal length 20 cm forms an image of an object placed 30 cm in front of it. Find image position and magnification.
\(f = -20\) cm, \(u = -30\) cm.
\[ \dfrac{1}{v} = \dfrac{1}{-20} + \dfrac{1}{-30} = -\dfrac{3+2}{60} = -\dfrac{1}{12} \]\(v = -12\) cm. Virtual, on the same side, 12 cm from the lens.
\(m = v/u = (-12)/(-30) = +0.4\). Erect, diminished (40% of object).
Example 6 — Power of a lens
Given: Find the power of a convex lens whose focal length is 25 cm. Also, find the focal length of a lens of power −2.5 D.
\(f = 25\ \text{cm} = 0.25\ \text{m}\). \(P = 1/f = 1/0.25 = +4\ \text{D}\).
For \(P = -2.5\ \text{D}\): \(f = 1/P = 1/(-2.5) = -0.4\ \text{m} = -40\ \text{cm}\). Negative f → concave lens of focal length 40 cm.
Example 7 — Combination of lenses
Given: A convex lens of focal length 20 cm and a concave lens of focal length 40 cm are placed in contact. Find the power and focal length of the combination.
\(P_1 = 1/0.20 = +5\ \text{D}\), \(P_2 = 1/(-0.40) = -2.5\ \text{D}\).
\(P = P_1 + P_2 = +5 - 2.5 = +2.5\ \text{D}\).
\(f = 1/P = 1/2.5 = 0.4\ \text{m} = 40\ \text{cm}\) (convex).
Example 8 — Finding focal length from two data
Given: An object placed 40 cm in front of a lens produces a real image 20 cm on the other side. Calculate the focal length and power of the lens. Is it convex or concave?
\(u = -40\) cm, \(v = +20\) cm.
\[ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{20} - \dfrac{1}{-40} = \dfrac{1}{20} + \dfrac{1}{40} = \dfrac{2+1}{40} = \dfrac{3}{40} \]\(f = 40/3 \approx 13.3\) cm (positive) ⇒ convex lens.
\(P = 1/f = 1/0.133 \approx +7.5\ \text{D}\).
Activity 9.3 — Tracing the Path of a Ray Through a Glass SlabL3 Apply
Aim: To trace a light ray through a rectangular glass slab and verify that the emergent ray is parallel to the incident ray.
Materials: Rectangular glass slab, four all-pins, white paper on drawing board, protractor, scale, pencil.
Procedure:
- Fix the paper on the drawing board and place the slab on it. Draw its outline ABCD.
- Remove the slab. Draw a normal at any point E on AB and draw an incident ray making an angle \(i\) (say 30°) with the normal.
- Fix two pins P1 and P2 on the incident ray. Replace the slab on its outline.
- Look through the opposite face CD and fix pins P3, P4 so that they appear in a straight line with P1, P2.
- Remove the slab and join P3P4 — this is the emergent ray. Join E to the point of emergence to get the refracted ray inside the slab.
Predict: Measure \(i\), \(r\) and the angle of emergence \(e\). How do they compare?
Measurement shows that \(r < i\) (ray bends toward normal entering glass), \(e = i\) (emergent ray parallel to incident ray) and the emergent ray is displaced sideways — the lateral shift. Snell's law \(n = \sin i/\sin r\) gives the refractive index of the glass slab as roughly 1.5 for crown glass.
Interactive — Lens Formula & Power Calculator
Enter \(u\) (cm) and \(f\) (cm) with proper signs. See image distance, magnification and power.
u (cm):
f (cm):
h (cm):
Competency-Based Questions
An optometrist advises Rohan's grandfather that his reading spectacles have a power of +2.5 D and his driving glasses a power of −1.5 D. Rohan wonders which lens is convex, which is concave and what their focal lengths are.
Q1. Classify both lenses and calculate their focal lengths. L3 Apply
Reading glasses: \(P=+2.5\) D (positive) → convex. \(f = 1/2.5 = 0.40\) m = 40 cm. Driving glasses: \(P=-1.5\) D (negative) → concave. \(f = 1/(-1.5) = -0.67\) m ≈ −66.7 cm.
Q2. (MCQ) A ray of light travels from water (n = 1.33) into air. What happens to it? L2 Understand
(b) Bends away from normal. Light speeds up going from denser (water) to rarer (air), so it bends away from the normal.
Q3. A ray of light passes from water (n₁ = 1.33) into glass (n₂ = 1.50). If the angle of incidence is 30°, find the angle of refraction. L3 Apply
\(n_1 \sin i = n_2 \sin r\). \(1.33 × \sin 30° = 1.50 × \sin r\). \(\sin r = 0.665/1.5 = 0.443\). \(r ≈ 26.3°\). Ray bends toward normal (denser medium).
Q4. (True/False + justify) A convex lens can form a virtual image. L4 Analyse
True. When the object is placed between the optical centre O and F1 of a convex lens, the refracted rays diverge and only their backward extensions meet. The image is virtual, erect and enlarged — this is exactly how a magnifying glass works.
Q5. Two lenses of power +3 D and −5 D are placed in contact. Find the combined power and focal length. What kind of lens does the combination behave like? L3 Apply
\(P = P_1 + P_2 = 3 + (-5) = -2\) D. \(f = 1/P = -0.50\) m = −50 cm. Negative focal length ⇒ the combination behaves like a concave (diverging) lens of focal length 50 cm.
Assertion–Reason Questions
Options: (A) Both A & R true, R correctly explains A. (B) Both A & R true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Assertion (A): The emergent ray from a rectangular glass slab is parallel to the incident ray.
Reason (R): The two refractions at the parallel faces of the slab bend the ray by equal and opposite amounts.
(A) — Both true; R correctly explains A. The sideways shift remains (lateral displacement).
Assertion (A): A concave lens cannot form a real image of a real object.
Reason (R): A concave lens always diverges the refracted rays, so they never meet in front of the lens.
(A) — Both true; R correctly explains A. Only virtual images are possible.
Assertion (A): The power of a lens of focal length 50 cm is 2 D.
Reason (R): Power \(P = 1/f\) where \(f\) is in metres.
(A) — Both true; R correctly explains A. \(f = 0.50\) m, \(P = 1/0.50 = +2\) D.
Frequently Asked Questions — Refraction, Lenses & Power
What is refraction, lenses & power in Class 10 Science (CBSE board)?
Refraction, Lenses & Power is a key topic in NCERT Class 10 Science Chapter 9 — Light - Reflection and Refraction. It explains refraction of light, snell's law, lenses, lens formula and power of a lens. Core ideas covered include refraction, refractive index, Snell's law, lens. Mastering this subtopic is essential for scoring well in the CBSE Class 10 Science board exam because board papers repeatedly test these concepts through MCQs, short answers and long-answer questions. This part gives a complete, exam-ready explanation with activities, diagrams and competency-based practice aligned to NCERT.
Why is refraction important in NCERT Class 10 Science?
Refraction is important in NCERT Class 10 Science because it forms the foundation for understanding refraction, lenses & power in Chapter 9 — Light - Reflection and Refraction. Without a clear idea of refraction, students cannot answer higher-order CBSE board questions involving refractive index, Snell's law, lens. Board papers regularly include 2-mark and 3-mark questions on this concept, and competency-based questions often link refraction to real-life situations. Building clarity here pays off directly in board marks.
How is refraction, lenses & power tested in the Class 10 Science CBSE board exam?
The CBSE Class 10 Science board exam tests refraction, lenses & power through a mix of 1-mark MCQs, 2-mark short answers, 3-mark explanations with examples, 5-mark descriptive questions (often with diagrams or balanced equations) and 4-mark competency-based questions. Expect direct questions on refraction, refractive index, Snell's law and application-based questions drawn from NCERT activities. Students who follow NCERT thoroughly and practice this chapter's questions consistently score in the 90%+ range.
What are the key terms to remember for refraction, lenses & power in Class 10 Science?
The key terms to remember for refraction, lenses & power in NCERT Class 10 Science Chapter 9 are: refraction, refractive index, Snell's law, lens, convex lens, concave lens. Each of these concepts carries exam weightage and regularly appears in the CBSE board paper. Write clear one-line definitions of every term in your revision notes and revisit them before the exam. Linking these terms visually through a flowchart or concept map makes recall easier during the Class 10 Science board exam.
Is Refraction, Lenses & Power included in the Class 10 Science syllabus for 2025–26 CBSE board exam?
Yes, Refraction, Lenses & Power is a part of the NCERT Class 10 Science syllabus (2025–26) prescribed by CBSE. It falls under Chapter 9 — Light - Reflection and Refraction — and is examined in the annual board paper. The current syllabus retains the full treatment of refraction, refractive index, Snell's law as per the NCERT textbook. Because CBSE bases every board question on NCERT, studying this part thoroughly ensures complete syllabus coverage and guarantees marks from this chapter.
How should I prepare refraction, lenses & power for the CBSE Class 10 Science board exam?
Prepare refraction, lenses & power for the CBSE Class 10 Science board exam in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of refraction, refractive index, Snell's law. Second, solve every in-text question and end-of-chapter exercise — CBSE questions often come directly from NCERT. Third, practice competency-based and assertion-reason questions to sharpen reasoning. Write answers in the exam-style format (point-wise with diagrams) and time yourself. This method delivers confidence and full marks in the board exam.
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