--- title: "Light – Reflection and Refraction: CBSE Class 10 Science previous year questions" url: https://www.swavid.com/cbse/class-10/science/pyq/light-reflection-and-refraction --- # Light – Reflection and Refraction: CBSE Class 10 Science previous year questions 16 questions from CBSE Class 10 board papers, newest first, each with full working. ## CBSE Class 10 Science Question Paper 2026 (Set 31/1/1) with Solutions ### Question 30 *1 mark · MCQ* A convex lens of focal length $15\text{ cm}$, is forming a real image. If the size of image is same as the size of object, then position of object and position of image will be, respectively : - $-15\text{ cm}$ and $-15\text{ cm}$ from lens - $-15\text{ cm}$ and $+15\text{ cm}$ from lens - $-30\text{ cm}$ and $+30\text{ cm}$ from lens - $-30\text{ cm}$ and $-30\text{ cm}$ from lens **Solution** 1. For a convex lens, the size of the image is equal to the size of the object when the object is placed at $2F$, i.e., at a distance of $2f$ from the optical centre. 2. Given focal length $f = 15\text{ cm}$, the object is at $-30\text{ cm}$ and the real image is formed at $+30\text{ cm}$. 3. Therefore, the correct option is (c). **Answer:** (c) $-30\text{ cm}$ and $+30\text{ cm}$ from lens > Common mistake: Confusing object and image distances with sign conventions or choosing the focal point instead of $2F$. ### Question 33 *2 marks · Case-based* The given figure shows image formation by a lens. Analyse the figure and answer the following questions : (a) What is the type of lens used for image formation in the given ray diagram ? (b) If the real image is formed at a distance of $30\text{ cm}$ from the lens and the size of image is twice the size of the object, then where was the object placed ? **Part (a)** 1. Observe the ray diagram where a real and enlarged image is formed by the lens. 2. A convex lens can form a real and magnified image when the object is placed between $F$ and $2F$. Answer (a): Convex lens **Part (b)** 1. Given: Image distance $v = +30\text{ cm}$ (real image), magnification $m = \frac{h_i}{h_o} = -2$ (real image is inverted). 2. Formula for magnification of a lens: $m = \frac{v}{u}$. 3. Substitute the values: $-2 = \frac{30}{u}$, which gives $u = \frac{30}{-2} = -15\text{ cm}$. 4. Result: The object was placed at a distance of $15\text{ cm}$ in front of the lens. Answer (b): $15\text{ cm}$ in front of the lens **Answer:** Analyse the lens ray diagram and calculate object position. > Common mistake: Forgetting the negative sign for magnification in the case of a real, inverted image. ### Question 38 *4 marks · Case-based* Read the following passage and answer the questions given : Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image. (a) A convex lens of focal length $20\text{ cm}$ is used to form an image. If an object is placed at $40\text{ cm}$ from the lens, what will be the position and nature of image ? (b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens. (c) (i) A lens combination consists of a convex lens of focal length $30\text{ cm}$ and a concave lens of focal length $15\text{ cm}$ placed together. Find the equivalent focal length and power of this lens combination. **Part (a)** 1. Given: f = +20 cm, u = -40 cm. 2. Formula: 1/v - 1/u = 1/f. 3. Calculation: 1/v = 1/20 - 1/40 = 1/40. 4. Result: v = +40 cm, image is real, inverted, and same size as object. Answer (a): v = 40 cm, real and inverted. **Part (b)** 1. Diagram: Draw a concave lens with an object between optical centre and focus. 2. The rays diverge and appear to meet behind the object to form a virtual, erect, and diminished image. Answer (b): Virtual, erect, and diminished image. **Part (c)(i)** 1. P1 = 100/30 = +3.33 D, P2 = 100/-15 = -6.67 D. 2. P = P1 + P2 = 3.33 - 6.67 = -3.34 D. 3. f = 100/P = -29.94 cm (approx -30 cm). Answer (c)(i): f = -30 cm, P = -3.34 D. **Answer:** The image is real, inverted, and formed at 40 cm. > Common mistake: Ignoring sign conventions for focal length and object distance. ### Question 38 (OR) *2 marks · Case-based* Read the following passage and answer the questions given : Lenses can form different types of images depending upon their focal length and position of object... (c) (ii) Two lenses are placed in contact. One is a concave lens with focal length $2\text{ m}$ and the other is a convex lens with focal length $1.5\text{ m}$. What type of lens will the combination behave as (convex or concave) ? Give reason. **Part (c)(ii)** 1. P1 = 1/-2 = -0.5 D, P2 = 1/1.5 = +0.67 D. 2. P = -0.5 + 0.67 = +0.17 D. 3. Since the net power is positive, the combination behaves as a convex lens. Answer (c)(ii): Convex lens. **Answer:** The combination behaves as a convex lens. > Common mistake: Incorrectly calculating the sum of powers. ## CBSE Class 10 Science Question Paper 2025 (Set 31/1/1) with Solutions ### Question 13 *1 mark · MCQ* Mirror 'X' is used to concentrate sunlight in solar furnace and Mirror 'Y' is fitted on the side of the vehicle to see the traffic behind the driver. Which of the following statements are true for the two mirrors ? (i) The image formed by mirror 'X' is real, diminished and at its focus. (ii) The image formed by mirror 'Y' is virtual, diminished and erect. (iii) The image formed by mirror 'X' is virtual, diminished and erect. (iv) The image formed by mirror 'Y' is real, diminished and at its focus. - (i) and (ii) - (ii) and (iii) - (iii) and (iv) - (i) and (iv) **Solution** 1. Mirror 'X' is a concave mirror used in solar furnaces to concentrate sunlight at its focus, forming a real and diminished image. 2. Mirror 'Y' is a convex mirror fitted as a rear-view mirror in vehicles, which always forms a virtual, erect, and diminished image. **Answer:** (b) (ii) and (iii) > Common mistake: Confusing the properties of concave mirrors used for convergence with convex mirrors used for wider field of view. ### Question 24 *2 marks · Numerical* An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm. Use lens formula to find the position of the image formed in this case. **Solution** 1. Given: Object distance $u = -60\text{ cm}$, Focal length of concave lens $f = -30\text{ cm}$. 2. Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ 3. Substitution: $\frac{1}{v} - \frac{1}{-60} = \frac{1}{-30}$ 4. Working: $\frac{1}{v} = \frac{1}{-30} - \frac{1}{60} = \frac{-2 - 1}{60} = \frac{-3}{60} = \frac{-1}{20}$ 5. Result: $v = -20\text{ cm}$ **Answer:** The image is formed at a distance of 20 cm in front of the lens. > Common mistake: Sign convention errors for concave lens focal length and object distance. ### Question 31 *3 marks · Short answer* Draw ray diagrams to show the nature, position and relative size of the image formed by a convex mirror when the object is placed (i) at infinity and (ii) between infinity and pole P of the mirror. **Solution** 1. Diagram for object at infinity: Draw a convex mirror with principal axis, focus F, and center of curvature C behind the mirror. Draw parallel incident rays reflecting away such that they appear to diverge from focus F. State nature: virtual, erect, and point-sized at F. 2. Diagram for object between infinity and pole P: Draw a convex mirror with object AB in front of it. Draw a ray parallel to the principal axis that gets reflected diverging from F, and a ray directed towards the center of curvature C that reflects back along the same path. Produce the reflected rays backward to intersect between P and F behind the mirror. 3. State nature for object between infinity and pole: virtual, erect, and diminished, formed behind the mirror between P and F. **Answer:** A convex mirror always forms virtual, erect images behind the mirror: point-sized at focus F when at infinity, and diminished between P and F when placed anywhere in front. > Common mistake: Drawing diverging rays incorrectly or failing to show virtual rays meeting behind the mirror with dotted lines. ### Question 36 *5 marks · Long answer* (a) (i) The power of a lens 'X' is $-2.5\text{ D}$. Name the lens and determine its focal length in cm. For which eye defect of vision will an optician prescribe this type of lens as a corrective lens ? (ii) "The value of magnification 'm' for a lens is $-2$." Using new Cartesian Sign Convention and considering that an object is placed at a distance of $20\text{ cm}$ from the optical centre of this lens, state : (I) the nature of the image formed; (II) size of the image compared to the size of the object; (III) position of the image, and (IV) sign of the height of the image. (iii) The numerical values of the focal lengths of two lenses A and B are $10\text{ cm}$ and $20\text{ cm}$ respectively. Which one of the two will show higher degree of convergence/divergence ? Give reason to justify your answer. **Part (a)(i)** 1. The lens has a negative power, so lens X is a concave lens. 2. The focal length $f = \frac{1}{P} = \frac{1}{-2.5\text{ D}} = -0.4\text{ m} = -40\text{ cm}$. 3. This type of lens is prescribed for the correction of myopia or nearsightedness. Answer (a)(i): Concave lens, focal length $-40\text{ cm}$, prescribed for myopia. **Part (a)(ii)** 1. A negative sign of magnification ($m = -2$) indicates that the image formed is real and inverted. 2. Since the magnitude of $m$ is $2$ (greater than $1$), the size of the image is enlarged or magnified compared to the size of the object. 3. Using magnification formula $m = \frac{v}{u} = -2$, with $u = -20\text{ cm}$, we get $v = -2 \times (-20\text{ cm}) = +40\text{ cm}$, so the image is formed at a distance of $40\text{ cm}$ on the other side of the lens. 4. Since the image is real and inverted, the sign of the height of the image is negative. Answer (a)(ii): (I) Real and inverted, (II) Magnified, (III) At $40\text{ cm}$ on the other side of the lens, (IV) Negative. **Part (a)(iii)** 1. Power of a lens is inversely proportional to its focal length, given by $P = \frac{1}{f}$. 2. Lens A has a smaller focal length ($10\text{ cm}$) compared to lens B ($20\text{ cm}$), which means lens A has a higher power. 3. Therefore, lens A will show a higher degree of convergence or divergence. Answer (a)(iii): Lens A, because it has a shorter focal length and hence higher power. **Answer:** Lens X is a concave lens with focal length $-40\text{ cm}$ used for myopia. For magnification $-2$, the image is real, inverted, magnified to twice the object size, formed at $40\text{ cm}$ on the other side, and has a negative height. Lens A has a higher degree of convergence/divergence. > Common mistake: Confusing the sign of the focal length for a concave lens or misinterpreting the sign of magnification for real versus virtual images. ### Question 36 (OR) *5 marks · Long answer* (b) (i) Draw a ray diagram to show the refraction of a ray of light through a rectangular glass slab when it falls obliquely from air into glass. (ii) State Snell's law of refraction of light. (iii) Differentiate between the virtual images formed by a convex lens and a concave lens on the basis of : (I) object distance, and (II) magnification. **Part (b) (i)** 1. Draw a rectangular glass slab PQRS. 2. Incident ray strikes face PQ obliquely, bends towards the normal inside the slab as the refracted ray, and emerges out from face SR as the emergent ray parallel to the incident ray (lateral displacement). Answer (b) (i): Ray diagram showing refraction through glass slab. **Part (b) (ii)** 1. Snell's law of refraction: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for the light of a given colour and for the given pair of media. 2. Formula: $\frac{\sin i}{\sin r} = \text{constant} = n_{21}$. Answer (b) (ii): $\frac{\sin i}{\sin r} = \text{constant}$ **Part (b) (iii)** 1. (I) Object distance: For a convex lens, a virtual image is formed when the object is placed between the optical centre and the principal focus ($u < f$). For a concave lens, a virtual image is formed for any position of the object ($u$ can be anywhere in front of the lens). 2. (II) Magnification: For a convex lens, the magnification for a virtual image is greater than 1 ($m > 1$, enlarged). For a concave lens, the magnification for a virtual image is always less than 1 ($m < 1$, diminished). Answer (b) (iii): Differences based on object distance and magnification stated. **Answer:** Ray diagram description, Snell's law, and difference between virtual images of convex and concave lenses. > Common mistake: Writing that a concave lens can form a real image, or confusing the magnification values for virtual images of convex and concave lenses. ## CBSE Class 10 Science Question Paper 2024 (Set 31/1/1) with Solutions ### Question 12 *1 mark · MCQ* At what distance from a convex lens should an object be placed to get an image of the same size as that of the object on a screen ? - Beyond twice the focal length of the lens. - At the principal focus of the lens. - At twice the focal length of the lens. - Between the optical centre of the lens and its principal focus. **Solution** 1. A convex lens produces a real image of the same size as the object only when the object is placed at $2F$ (twice the focal length). 2. The image is also formed at $2F$ on the other side of the lens. **Answer:** (c) At twice the focal length of the lens. > Common mistake: Confusing the condition for same-sized image with the focus or optical centre positions. ### Question 24 *2 marks · Numerical* An object is placed at a distance of $10\text{ cm}$ from a convex mirror of focal length $15\text{ cm}$. Find the position of the image formed by the mirror. **Solution** 1. Given: Object distance $u = -10\text{ cm}$, Focal length $f = +15\text{ cm}$ 2. Formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ 3. Substitution: $\frac{1}{15} = \frac{1}{v} + \frac{1}{-10}$ 4. Working: $\frac{1}{v} = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$ 5. Result: $v = +6\text{ cm}$ **Answer:** $+6\text{ cm}$ > Common mistake: Incorrect sign convention for $u$ and $f$ in a convex mirror. ### Question 39 *4 marks · Case-based* Study the data given below showing the focal length of three concave mirrors A, B and C and the respective distances of objects placed in front of the mirrors : (i) In which one of the above cases the mirror will form a diminished image of the object ? Justify your answer. (ii) List two properties of the image formed in case 2. (iii) (A) What is the nature and size of the image formed by mirror C ? Draw ray diagram to justify your answer. **Part (i)** 1. In Case 1, the object distance ($u = -45\text{ cm}$) is greater than the radius of curvature ($2f = -40\text{ cm}$), so the object is placed beyond C. 2. When an object is placed beyond C in a concave mirror, the image formed is real, inverted, and diminished. Answer (i): Case 1, because the object is placed beyond the center of curvature. **Part (ii)** 1. In Case 2, the object is at $u = -30\text{ cm}$ and $f = -15\text{ cm}$, meaning the object is at the center of curvature C ($u = 2f$). 2. The image formed is real, inverted, and of the same size as the object. Answer (ii): The image is real and inverted (or equal in size to the object). **Part (iii) (A)** 1. In Case 3, the object distance $u = -20\text{ cm}$ is less than the focal length $f = -30\text{ cm}$, so the object lies between the pole and the principal focus. 2. The nature of the image formed is virtual and erect, and its size is enlarged. 3. Diagram: Draw a concave mirror with object between P and F, showing two rays (one parallel to principal axis passing through focus, another passing through center of curvature) meeting behind the mirror to form a virtual, erect, and enlarged image. Answer (iii) (A): Virtual, erect, and enlarged image. (Refer to standard textbook ray diagram for object between P and F of a concave mirror). **Answer:** Case 1 forms a diminished image, case 2 forms a real and inverted image, and case 3 forms an enlarged virtual image. > Common mistake: Incorrectly identifying object positions with respect to focal length and radius of curvature. ### Question 39 (OR) *2 marks · Numerical* (iii) (B) An object is placed at a distance of $18\text{ cm}$ from the pole of a concave mirror of focal length $12\text{ cm}$. Find the position of the image formed in this case. **Solution** 1. Given: Object distance $u = -18\text{ cm}$, Focal length $f = -12\text{ cm}$. 2. Formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ 3. Substitution: $\frac{1}{-12} = \frac{1}{v} + \frac{1}{-18}$ 4. Working: $\frac{1}{v} = -\frac{1}{12} + \frac{1}{18} = \frac{-3 + 2}{36} = -\frac{1}{36}$ 5. Result: $v = -36\text{ cm}$ **Answer:** The image is formed at a distance of $36\text{ cm}$ in front of the concave mirror. > Common mistake: Sign convention errors for object distance and focal length of a concave mirror. ## CBSE Class 10 Science Question Paper 2023 (Set 31/1/1) with Solutions ### Question 30 *3 marks · Short answer* (a) Complete the following ray diagram to show the formation of image : [Diagram showing concave mirror with object AB and rays] (b) Mention the nature, position and size of the image formed in this case. (c) State the sign of the image distance in this case using the Cartesian sign convention. **Solution** 1. A ray parallel to the principal axis passes through the principal focus after reflection. 2. A ray passing through the focus reflects parallel to the principal axis, forming a real, inverted and enlarged image beyond the centre of curvature when the object is placed between F and C. **Answer:** Nature: Real and inverted; Position: Beyond C; Size: Enlarged; Sign of image distance: Negative. > Common mistake: Forgetting that image distance for a real image formed by a concave mirror is negative. ### Question 39 *4 marks · Case-based* Hold a concave mirror in your hand and direct its reflecting surface towards the sun. Direct the light reflected by the mirror on to a white card-board held close to the mirror. Move the card-board back and forth gradually until you find a bright, sharp spot of light on the board. This spot of light is the image of the sun on the sheet of paper; which is also termed as "Principal Focus" of the concave mirror. [Diagram showing parallel rays converging at principal focus of concave mirror] (a) List two applications of concave mirror. (b) If the distance between the mirror and the principal focus is $15\text{ cm}$, find the radius of curvature of the mirror. (c) Draw a ray diagram to show the type of image formed when an object is placed between pole and focus of a concave mirror. **Part (a)** 1. Application 1: Used in torches, search-lights and vehicle headlights to get powerful parallel beams of light. 2. Application 2: Used as shaving mirrors to see a larger image of the face. Answer (a): Used in torches and as shaving mirrors **Part (b)** 1. Focal length ($f$) = $15\text{ cm}$ 2. Radius of curvature ($R$) = $2f = 2 \times 15\text{ cm} = 30\text{ cm}$ Answer (b): 30 cm **Part (c)** 1. Diagram: Draw a concave mirror with pole P, focus F, and centre of curvature C. 2. Place object between P and F. Draw one ray parallel to principal axis passing through F, and another ray through C. 3. Produce rays backward to intersect behind the mirror, forming an enlarged, virtual and erect image. Answer (c): Ray diagram showing virtual, erect, and magnified image behind the mirror. **Answer:** Concave mirrors are used in torches and shaving mirrors, radius of curvature is 30 cm, and ray diagram shows a virtual and erect image. > Common mistake: Forgetting arrows on light rays in ray diagrams. ### Question 39 (OR) *2 marks · Case-based* An object $10\text{ cm}$ in size is placed at $100\text{ cm}$ in front of a concave mirror. If its image is formed at the same point where the object is located, find : (i) focal length of the mirror, and (ii) magnification of the image formed with sign as per Cartesian sign convention. **Part (i)** 1. Given object size $h = 10\text{ cm}$, object distance $u = -100\text{ cm}$. 2. Since the image is formed at the same point where the object is located, the object is placed at the centre of curvature C. 3. Therefore, image distance $v = -100\text{ cm}$ and radius of curvature $R = -100\text{ cm}$. 4. Focal length $f = \frac{R}{2} = \frac{-100\text{ cm}}{2} = -50\text{ cm}$. Answer (i): -50 cm **Part (ii)** 1. Magnification $m = -\frac{v}{u}$ 2. Substitute values: $m = -\frac{-100\text{ cm}}{-100\text{ cm}} = -1$ Answer (ii): -1 **Answer:** Focal length is -50 cm and magnification is -1. > Common mistake: Writing positive sign for magnification of a real and inverted image formed at the centre of curvature. ## Related pages - [Light – Reflection and Refraction: NCERT solutions](https://www.swavid.com/science/class/10/chapter/light-reflection-and-refraction/ncert-solutions) - [All CBSE Class 10 Science papers](https://www.swavid.com/cbse/class-10/science/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student