Class 9 Science Chapter 11 Light Important Questions with CBSE Board-Pattern Answers

Light is not just a standalone chapter—it bridges Class 9 fundamentals and Class 10 advanced optics. In the 2024-25 rationalized curriculum, examiners focus on three high-weightage areas: (1) Laws of Reflection and plane mirror image properties; (2) Concave and convex mirror numerical problems (radius, focal length, magnification); and (3) Lens function and dispersion of white light. Board examiners typically set 1–2 questions from mirror ray diagrams, 1 numerical on magnification or focal length, and 1 short-answer on refraction or rainbow. The expected question split is: 1-mark (1 question), 2-mark (1–2 questions), 3-mark (1 question), and 5-mark (occasional numericals). Mastering the patterns in this guide ensures you score full marks on the predictable portions while gaining confidence for unseen variations.

**Question 1:** A light ray hits a plane mirror at an angle of 30° to the normal. The angle of reflection is: (a) 30° (b) 60° (c) 90° (d) 120° **Answer:** (a) 30° **Explanation:** The law of reflection states that the angle of incidence equals the angle of reflection, both measured from the normal. If the incident ray makes 30° with the normal, the reflected ray also makes 30° with the normal. **Question 2:** A concave mirror has a focal length of 10 cm. Its radius of curvature is: (a) 5 cm (b) 10 cm (c) 20 cm (d) 40 cm **Answer:** (c) 20 cm **Explanation:** The relationship between radius (R) and focal length (f) is R = 2f. If f = 10 cm, then R = 2 × 10 = 20 cm. **Question 3:** When white light passes through a prism, it splits into seven colours. This phenomenon is called: (a) Refraction (b) Dispersion (c) Reflection (d) Diffraction **Answer:** (b) Dispersion **Explanation:** Dispersion is the splitting of white light into its constituent colours (VIBGYOR) due to different wavelengths refracting at slightly different angles in the prism. **Question 4:** A convex lens is also known as a: (a) Diverging lens (b) Converging lens (c) Plane lens (d) Cylindrical lens **Answer:** (b) Converging lens **Explanation:** Convex lenses are thicker at the centre and converge parallel rays of light to a focal point. They are used in magnifying glasses and camera lenses. **Question 5:** An image formed by a plane mirror is always: (a) Real, inverted, same size (b) Virtual, erect, same size (c) Real, erect, larger (d) Virtual, inverted, smaller **Answer:** (b) Virtual, erect, same size **Explanation:** Plane mirrors always form virtual images (behind the mirror) that are upright and the same size as the object. This is why you see yourself life-sized in a bathroom mirror.

**Question 1:** State the laws of reflection of light. **Answer:** The two laws of reflection are: (1) The incident ray, reflected ray, and normal all lie in the same plane. (2) The angle of incidence is equal to the angle of reflection, where both angles are measured from the normal to the mirror surface. **Worked Example:** If light strikes a mirror at 40° to the normal, it reflects at 40° on the other side of the normal. **Question 2:** How does a concave mirror form a real, inverted, magnified image? **Answer:** When an object is placed between the centre of curvature (C) and the focal point (F) of a concave mirror, the reflected rays converge to form a real, inverted image. The image is formed beyond the centre of curvature, making it larger than the object. This property is used in shaving mirrors and dental mirrors, where magnification is desired. **Question 3:** What is the difference between a convex lens and a concave lens? **Answer:** | Property | Convex Lens | Concave Lens | |----------|-------------|---------------| | Shape | Thick at centre, thin at edges | Thin at centre, thick at edges | | Effect on light | Converges parallel rays | Diverges parallel rays | | Type | Converging | Diverging | | Focal length | Positive | Negative | | Common use | Magnifying glass, camera | Peephole, eyeglass for myopia | **Question 4:** Define lateral magnification and write its formula for mirrors. **Answer:** Lateral magnification (m) is the ratio of the height of the image to the height of the object. For mirrors: m = h' / h = −v / u where h' = image height, h = object height, v = image distance, u = object distance. Negative magnification indicates an inverted image; positive indicates an erect image. Example: If an object 2 cm tall forms an image 6 cm tall, m = 6/2 = 3 (magnified, erect). **Question 5:** Why does a rainbow form? Name the phenomenon involved. **Answer:** A rainbow forms when sunlight enters water droplets in the atmosphere at a certain angle (typically 42° from the antisolar point). Inside each droplet, white light undergoes refraction (bending), dispersion (splitting into colours), and internal reflection. Red light bends the least, appearing at the outer edge; violet bends the most, appearing at the inner edge. The phenomenon is called dispersion combined with refraction and reflection.

**Question 1:** An object 4 cm tall is placed 30 cm in front of a concave mirror of focal length 10 cm. Calculate (a) image distance, (b) magnification, and (c) describe the nature of the image. **Solution:** Given: h = 4 cm, u = 30 cm, f = 10 cm Using mirror formula: 1/f = 1/u + 1/v 1/10 = 1/30 + 1/v 1/v = 1/10 − 1/30 = 3/30 − 1/30 = 2/30 = 1/15 v = 15 cm (a) **Image distance:** 15 cm (in front of mirror) (b) **Magnification:** m = −v/u = −15/30 = −0.5 Image height = m × h = −0.5 × 4 = −2 cm (c) **Nature:** Real, inverted, diminished (half the size), formed between F and C. **Question 2:** An object is placed 60 cm from a convex mirror of focal length 20 cm. Find the image distance and magnification. **Solution:** Given: u = 60 cm, f = +20 cm (convex, so f is positive) Using: 1/f = 1/u + 1/v 1/20 = 1/60 + 1/v 1/v = 1/20 − 1/60 = 3/60 − 1/60 = 2/60 = 1/30 v = 30 cm (behind the mirror, virtual) **Magnification:** m = −v/u = −(−30)/60 = +0.5 **Conclusion:** Image is virtual, erect, diminished to half size, located 30 cm behind the mirror. **Question 3:** A convex lens has a focal length of 20 cm. An object is placed 30 cm in front of the lens. Find (a) image distance, (b) magnification, and (c) nature of image. **Solution:** Given: f = 20 cm, u = 30 cm Using lens formula: 1/f = 1/u + 1/v 1/20 = 1/30 + 1/v 1/v = 1/20 − 1/30 = 3/60 − 2/60 = 1/60 v = 60 cm (a) **Image distance:** 60 cm (on the opposite side of the object) (b) **Magnification:** m = −v/u = −60/30 = −2 Image height = −2 × object height (inverted, doubled) (c) **Nature:** Real, inverted, magnified (2× larger). **Question 4:** Explain with a ray diagram how a prism disperses white light into a spectrum. **Answer:** When white light enters a prism: 1. **Refraction at first surface:** Light slows down and bends toward the normal as it enters the denser glass. 2. **Dispersion inside prism:** Different colours have different speeds in glass. Red (longest wavelength) bends least; violet (shortest) bends most. 3. **Refraction at second surface:** Light speeds up and bends away from the normal as it exits into air. 4. **Result:** Colours separate into a spectrum (VIBGYOR), with red on top and violet at the bottom. [Ray diagram shows incoming white ray splitting into seven coloured rays, with violet bent most and red least]

**Question 1:** Explain the formation of images by concave mirrors for different object positions. Draw ray diagrams for at least two cases. **Complete Answer:** A concave mirror forms different types of images depending on the object's position relative to the focal point (F) and centre of curvature (C). The relationship R = 2f always holds (R = radius, f = focal length). **Case 1: Object beyond C (u > 2f)** - Image formed between F and C - Image distance: F < v < 2f - Nature: Real, inverted, diminished (smaller than object) - Uses: Astronomical telescopes - Ray diagram: Object beyond C → converging rays below the principal axis → image between F and C **Case 2: Object at C (u = 2f)** - Image formed at C - Image distance: v = 2f - Magnification: m = −1 (same size) - Nature: Real, inverted, equal size - Ray diagram: Object at C → rays converge at C → image at C **Case 3: Object between C and F (F < u < 2f)** - Image formed beyond C - Image distance: v > 2f - Nature: Real, inverted, magnified (larger than object) - Uses: Shaving mirrors, dentist's mirrors - Ray diagram: Object between C and F → rays diverge after reflection → meet beyond C **Case 4: Object at F (u = f)** - Image formed at infinity - Image distance: v = ∞ - Nature: Real, highly magnified - Ray diagram: Object at F → reflected rays parallel to principal axis → no image formation (or image at ∞) **Case 5: Object between F and P (u < f)** [P = pole/vertex] - Image formed behind the mirror - Image distance: v is negative (virtual) - Nature: Virtual, erect, magnified - Uses: Shaving mirrors (when held very close to face) - Ray diagram: Object between F and P → reflected rays diverge → appear to come from behind mirror **Magnification formula:** m = −v/u. A negative m means inverted; positive means erect. --- **Question 2:** What is dispersion of light? How does it explain the formation of a rainbow? Why do we see VIBGYOR in that order? **Complete Answer:** **Dispersion Definition:** Dispersion is the phenomenon in which white light splits into its seven constituent colours (VIBGYOR = Violet, Indigo, Blue, Green, Yellow, Orange, Red) when passing through a prism or water droplet. Each colour has a different wavelength (λ), and materials refract different wavelengths at slightly different angles. **Rainbow Formation (Detailed Process):** 1. **Condition:** Sunlight must enter water droplets at ≈ 42° from the antisolar point (the point directly opposite the sun from the observer's perspective). 2. **Refraction at entry:** White sunlight enters a spherical water droplet. The refractive index of water (n ≈ 1.33) causes bending toward the normal: - Red bends least (≈ 40°) - Violet bends most (≈ 42°) 3. **Dispersion inside droplet:** As light travels through the droplet, colours separate due to varying speeds. Red travels fastest; violet slowest. 4. **Internal reflection:** Light reflects off the back inner surface of the droplet (total internal reflection at ≈ 75°). 5. **Refraction at exit:** As light exits the droplet, it bends away from the normal again, widening the angle between colours. 6. **Final angle:** The light exits at approximately 42° from the antisolar point. A secondary rainbow (fainter, colours reversed) forms if light undergoes two internal reflections (≈ 50° from antisolar point). **Why VIBGYOR Order?** The refractive index n varies slightly with wavelength (chromatic dispersion). Violet light has the shortest wavelength (λ ≈ 400 nm) and highest refractive index, so it bends most—appearing at the inner, darker edge of the rainbow. Red has the longest wavelength (λ ≈ 700 nm) and lowest refractive index, so it bends least—appearing at the outer edge. **Mathematical relationship:** Snell's law: n₁ sin θ₁ = n₂ sin θ₂ For violet: n_violet is larger → θ₂ is larger → more bending → inner edge For red: n_red is smaller → θ₂ is smaller → less bending → outer edge **Why rainbows only appear at ≈ 42°:** This is the angle of minimum deviation for light in water droplets. Light at other angles either escapes or doesn't converge toward the observer's eye. --- **Question 3:** State and derive the lens formula. Solve a practical numerical using this formula. **Complete Answer:** **Lens Formula (Derivation):** For a thin lens, the relationship between object distance (u), image distance (v), and focal length (f) is derived using geometry and the refraction of parallel rays: **1/f = 1/u + 1/v** **Derivation (outline):** - Consider a ray from object entering the lens at height h from principal axis - Parallel ray bends toward focal point by angle θ (small angle approximation) - Using similar triangles from object, lens, and focal point positions - Combining equations for ray paths → 1/f = 1/u + 1/v **Sign Convention:** - Object distance (u): Always negative for real objects (in front of lens) - Image distance (v): Positive if real image (behind lens for convex), negative if virtual (same side as object) - Focal length (f): Positive for converging (convex) lens, negative for diverging (concave) lens **Practical Numerical:** A concave lens of focal length −15 cm forms an image 10 cm in front of the lens. Find the object distance. **Given:** f = −15 cm, v = −10 cm (virtual image, same side as object) **Using:** 1/f = 1/u + 1/v 1/(−15) = 1/u + 1/(−10) −1/15 = 1/u − 1/10 1/u = −1/15 + 1/10 = −2/30 + 3/30 = 1/30 u = 30 cm **Magnification:** m = −v/u = −(−10)/30 = +10/30 = +1/3 Image is virtual, erect, and diminished to 1/3 the object height. **Verification:** A concave lens always produces virtual, erect, diminished images, which matches our result. ✓

**Case-Study Question:** A photographer is using a concave mirror of focal length 20 cm to focus light from a distant landscape onto a screen to create a real image. The mirror's radius of curvature is 40 cm. However, she observes that when she moves the screen closer to the mirror, the image becomes larger but blurrier. When she moves it further, the image shrinks and sharpens again. Later, she switches to a convex mirror for a rearview camera on a car. **(a) Explain why the image size changes when the screen position changes.** **Answer:** The magnification m = −v/u depends directly on the image distance (v). As the screen (image position) moves: - If v increases (screen further from mirror), magnification increases, image becomes larger - If v decreases (screen closer to mirror), magnification decreases, image becomes smaller The "blurriness" occurs because different rays don't converge exactly at the screen position—this is spherical aberration. When v matches the calculated position from the lens formula (1/f = 1/u + 1/v), all rays converge perfectly, and the image is sharp. **(b) Why is the concave mirror not suitable for a car rearview, but a convex mirror is?** **Answer:** Concave mirrors have a limited field of view because they produce real, inverted, magnified images only for objects beyond the focal point. A driver needs to see a wide area behind the car. A convex mirror, however: - Produces virtual, erect, diminished images for all object distances - Has a much wider field of view (up to 180°) - Shows a wider area of the road in a compact mirror - Allows the driver to check blind spots safely Example: A car 5 m behind appears smaller in a convex mirror (m = +0.25), but the driver can see it plus vehicles in adjacent lanes—impossible with a concave mirror. **(c) Calculate the screen distance needed to project a 2 cm tall landscape image using the given concave mirror, if the distant landscape acts as an object at approximately 10 m (1000 cm).** **Solution:** Given: f = 20 cm, u ≈ 1000 cm (very large, distant object) Using: 1/f = 1/u + 1/v 1/20 = 1/1000 + 1/v 1/v = 1/20 − 1/1000 = 50/1000 − 1/1000 = 49/1000 v ≈ 1000/49 ≈ 20.4 cm **Screen distance ≈ 20.4 cm** (just beyond the focal point) **Magnification:** m = −v/u = −(20.4)/(1000) ≈ −0.0204 Image height = 0.0204 × object height For the image to be 2 cm, the landscape object height must be ≈ 2/0.0204 ≈ **98 cm** tall. This demonstrates why distant landscapes produce tiny inverted images on the screen—suitable for projecting onto a screen in a camera obscura setup.

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