Class 9 Physics Chapter 9: Ray Optics and Optical Instruments Important Questions with Solutions

Ray Optics and Optical Instruments accounts for approximately 8–10% of the total Physics paper in CBSE Class 9 finals, translating to 8–12 marks in theory and practicals combined. The chapter bridges theoretical optics (laws of reflection and refraction) with practical applications (microscopes, telescopes, corrective lenses). Board examiners focus on: (1) Numerical problem-solving using lens formula and magnification; (2) Ray diagram construction for object placement at different distances from a lens; (3) Understanding how optical instruments magnify or diminish images; (4) Real-world applications like myopia correction and astronomical observations. Recent board papers show a marked shift toward application-based questions and HOTS (Higher Order Thinking Skills) questions that ask students to analyse optical phenomena in everyday life—for example, why a concave mirror is used in searchlights, or how a simple microscope magnifies. Multiple-choice questions often test conceptual clarity on critical angles, refractive indices, and focal length relationships. By practising diverse question patterns, you not only secure marks but also develop the analytical thinking required for Class 10 and competitive exams.

**Q1. The refractive index of glass is 1.5. The critical angle for light travelling from glass to air is:** (A) 30° (B) 42° (C) 45° (D) 60° **Answer:** (B) 42° **Explanation:** Critical angle θc = sin⁻¹(1/n) = sin⁻¹(1/1.5) ≈ 41.8° ≈ 42°. When light travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle, total internal reflection occurs. **Q2. A convex lens of focal length 10 cm is used as a magnifying glass. The object should be placed:** (A) At the centre of curvature (B) Between the pole and focal point (C) At infinity (D) At the focal point **Answer:** (B) Between the pole and focal point **Explanation:** For a convex lens to act as a magnifying glass (virtual, erect, and magnified image), the object must lie between the optical centre and the focal point, i.e., u < f. **Q3. A plane mirror always produces:** (A) A real and inverted image (B) A virtual and erect image (C) A real and erect image (D) A virtual and inverted image **Answer:** (B) A virtual and erect image **Explanation:** Plane mirrors form images that are virtual (located behind the mirror), erect (same orientation as object), laterally inverted, and equidistant from the mirror as the object. **Q4. The lens used in a microscope for initial magnification is called:** (A) Eyepiece (B) Objective lens (C) Condenser (D) Field lens **Answer:** (B) Objective lens **Explanation:** The objective lens is placed closest to the specimen and produces a real, inverted, and magnified image. The eyepiece then magnifies this intermediate image further. **Q5. When light is incident normally on a surface, the angle of incidence is:** (A) 0° (B) 30° (C) 45° (D) 90° **Answer:** (A) 0° **Explanation:** When a ray strikes a surface perpendicularly (along the normal), the angle between the ray and the normal is 0°, so angle of incidence = 0°.

**Q1. Define the term 'focal length' of a lens. Write the lens formula.** **Answer:** Focal length is the distance from the optical centre of a lens to its focal point (F), where parallel rays converge (for a convex lens) or appear to diverge (for a concave lens). Focal length is denoted by f and is measured in centimetres or metres. Lens formula: **1/f = 1/u + 1/v**, where u is the object distance, v is the image distance, and f is the focal length. **Q2. Explain the law of reflection with a neat diagram.** **Answer:** Law of Reflection states: (1) The incident ray, reflected ray, and normal all lie in the same plane. (2) The angle of incidence equals the angle of reflection (i = r). [A diagram should show a mirror surface with incident ray, reflected ray, normal (perpendicular to mirror), angle i on the left side of normal, and angle r on the right side of normal, both marked equal.] **Q3. A concave mirror has a radius of curvature of 20 cm. Calculate its focal length.** **Answer:** Focal length f = R/2, where R is the radius of curvature. Given: R = 20 cm f = 20/2 = **10 cm** The focal length of the concave mirror is 10 cm, located on the same side as the object. **Q4. Distinguish between real and virtual images in one sentence each.** **Answer:** **Real image:** An image formed by the actual convergence of light rays at a point; it is inverted, can be projected on a screen, and is formed in front of a concave mirror or by a convex lens when object is beyond the focal point. **Virtual image:** An image formed by the apparent divergence of light rays (they appear to come from a point); it is erect, cannot be projected on a screen, and is formed by a convex mirror or plane mirror. **Q5. What is meant by 'refraction of light'? Give one example.** **Answer:** Refraction is the bending of light rays when they pass from one transparent medium to another of different optical density. The ray bends toward the normal if entering a denser medium, and away from the normal if entering a rarer medium. **Example:** A pencil immersed in water appears bent at the water–air interface because light rays from the submerged portion refract as they exit the water.

**Q1. An object of height 3 cm is placed at a distance of 15 cm from a convex lens of focal length 10 cm. Find the position, nature, and size of the image.** **Solution:** Given: h₀ = 3 cm, u = 15 cm, f = 10 cm Using lens formula: 1/f = 1/u + 1/v 1/10 = 1/15 + 1/v 1/v = 1/10 − 1/15 = (3 − 2)/30 = 1/30 v = 30 cm Image position: **30 cm on the opposite side of the lens (real side).** Magnification m = v/u = 30/15 = 2 Image height h₁ = m × h₀ = 2 × 3 = **6 cm** Nature: Real, inverted, magnified (m > 1) **Q2. Draw a ray diagram for a concave mirror when the object is placed at the centre of curvature (C). State the nature and position of the image.** **Solution:** [Ray diagram should show: (1) A concave mirror surface; (2) Points P (pole), F (focal point), and C (centre of curvature) marked on the principal axis; (3) Object AB placed at C; (4) Two rays—one parallel to principal axis reflecting through F, and one through C reflecting back along the same path; (5) Image A'B' formed at C.] Nature of image: Real, inverted, same size as object (m = −1) Position: At the centre of curvature (C), on the object side This property is used in shaving mirrors where the face at a certain distance produces a magnified real image. **Q3. Explain why a concave lens always forms a virtual image, irrespective of object position.** **Solution:** A concave lens is a diverging lens—it causes parallel rays to spread out as if emanating from the focal point on the object side. For a concave lens, the focal length f is negative. Using lens formula: 1/f = 1/u + 1/v For any positive object distance u: 1/v = 1/f − 1/u Since f is negative and 1/u is positive, 1/v is always negative, making v negative. A negative image distance means the image forms on the same side as the object—this is the definition of a virtual image. The rays diverge and never actually meet; they appear to diverge from a point behind the lens. Additionally, magnification m = v/u is always less than 1 (and positive), so the image is always erect and diminished. **Q4. A ray of light travels from glass (n = 1.5) to air (n = 1). If the angle of incidence is 30°, calculate the angle of refraction. Will total internal reflection occur if the angle of incidence is increased to 50°?** **Solution:** Using Snell's Law: n₁ sin θ₁ = n₂ sin θ₂ 1.5 × sin 30° = 1 × sin θ₂ 1.5 × 0.5 = sin θ₂ sin θ₂ = 0.75 θ₂ ≈ **48.6°** Critical angle θc = sin⁻¹(n₂/n₁) = sin⁻¹(1/1.5) ≈ 41.8° When angle of incidence = 50°: Since 50° > 41.8° (critical angle), **yes, total internal reflection will occur**. The light ray will reflect back into the glass medium instead of refracting into air.

**Q1. Describe the structure and working principle of a simple microscope (magnifying glass). Derive the expression for magnification.** **Full Solution:** **Structure:** A simple microscope consists of a single convex lens of short focal length (typically 10–15 mm) mounted in a frame with an eyepiece to allow comfortable viewing. **Working Principle:** The object to be magnified is placed at a distance slightly less than the focal length (u < f) from the convex lens. This arrangement produces a virtual, erect, and highly magnified image on the same side of the lens. The magnified image is formed at the least distance of distinct vision (D = 25 cm for the standard human eye) or at infinity for relaxed viewing. **Magnification Formula (for image at least distance of distinct vision):** When the final image forms at distance D from the eye (typically 25 cm): Using lens formula: 1/f = 1/u + 1/v For a virtual image (v negative): 1/f = 1/u − 1/|v| Magnification M = (1 + D/f) where D is the least distance of distinct vision (25 cm). **When image is at infinity (for relaxed viewing):** M = D/f **Example:** For a lens with f = 5 cm, M = 25/5 = 5×, meaning the image is 5 times magnified. **Key Points:** • Single convex lens used • Virtual, erect, magnified image • Used for examining small objects like insects, coins, stamps • Magnification typically ranges from 5× to 20× **Q2. An object is placed at various distances from a concave mirror of focal length 15 cm. Draw ray diagrams and describe the image formation in each case:** (a) u > 2f (b) u = 2f (c) f < u < 2f (d) u < f (e) u = f **Full Solution:** [Five separate ray diagrams should be provided for each case. Key features for each:] **(a) u > 2f (Object beyond centre of curvature):** - Image position: f < v < 2f - Nature: Real, inverted, diminished (m < 1) - Ray diagram: Two rays—one parallel reflecting through F, one through C reflecting back—meet between C and F **(b) u = 2f (Object at centre of curvature):** - Image position: v = 2f (at C) - Nature: Real, inverted, same size (m = −1) - Ray diagram: Both rays reflect and meet exactly at C on the object side **(c) f < u < 2f (Object between F and C):** - Image position: v > 2f (beyond C) - Nature: Real, inverted, magnified (m > 1) - Ray diagram: Rays meet far from the mirror on the object side **(d) u < f (Object between P and F):** - Image position: v is negative (virtual), on same side as object - Nature: Virtual, erect, magnified (m > 1) - Ray diagram: Rays diverge; when extended backward, they meet behind the mirror—used in shaving mirrors **(e) u = f (Object at focal point):** - Image position: v = ∞ (at infinity) - Nature: Real, inverted, highly magnified (m = ∞) - Ray diagram: Reflected rays are parallel to principal axis and never meet **Q3. Explain the phenomenon of total internal reflection. State the conditions necessary for it to occur and derive the expression for critical angle. Give two real-world applications.** **Full Solution:** **Definition:** Total internal reflection is the phenomenon in which light travelling from a denser medium (higher refractive index) to a rarer medium (lower refractive index) is completely reflected back into the denser medium when the angle of incidence exceeds a certain critical angle. **Conditions for Total Internal Reflection:** 1. Light must travel from a denser to a rarer medium (n₁ > n₂) 2. Angle of incidence must be greater than or equal to the critical angle (i ≥ θc) 3. The interface must be smooth **Derivation of Critical Angle Expression:** At the critical angle θc, the refracted ray travels along the interface, so the angle of refraction = 90°. Applying Snell's Law: n₁ sin θc = n₂ sin 90° n₁ sin θc = n₂ × 1 sin θc = n₂/n₁ **θc = sin⁻¹(n₂/n₁)** **Example:** For glass (n = 1.5) to air (n = 1): θc = sin⁻¹(1/1.5) ≈ 41.8° If light hits the glass–air interface at > 41.8°, it reflects back into the glass. **Real-World Applications:** 1. **Optical Fibres:** Signals travel through hair-thin glass fibres via total internal reflection, enabling high-speed internet and telecommunications. The fibre's refractive index is higher than air, so light bounces internally along the entire length without signal loss. 2. **Diamond Sparkle:** Diamonds have a very high refractive index (n ≈ 2.42). When light enters a cut diamond, it undergoes total internal reflection multiple times before exiting. This traps and disperses light brilliantly, giving diamonds their characteristic sparkle. Other applications: Prism binoculars, periscopes, and emergency reflection straps for safety wear.

**Q. Telescope vs. Microscope—A Comparative Analysis** A student observes that both a telescope and a microscope use two convex lenses, yet they serve opposite purposes: one magnifies distant celestial objects, and the other magnifies tiny nearby objects. The student wonders: How can the same type of lens configuration produce such different results? **Given information:** - Simple telescope focal lengths: objective f₀ = 100 cm, eyepiece fe = 5 cm - Simple microscope focal lengths: objective f₀ = 1 cm, eyepiece fe = 5 cm **Questions:** (a) **Structural difference:** How do the focal lengths of the objective and eyepiece in a telescope differ from those in a microscope? Explain why this difference matters. (3 marks) (b) **Magnification:** Calculate the magnification of each instrument. What does this tell you about the relationship between focal length and magnification? (3 marks) (c) **Object placement:** In a telescope, where is the object relative to the objective? In a microscope, where is the object relative to the objective? Why must these positions differ? (2 marks) (d) **Real-world constraint:** A student constructs a homemade microscope using a double-convex lens of focal length 0.5 cm and an eyepiece of focal length 5 cm. However, she complains that the magnified image appears dim. Explain why and suggest a solution. (2 marks) **Model Solution:** **(a) Focal length differences:** In a **telescope**: The objective has a large focal length (f₀ = 100 cm), and the eyepiece has a short focal length (fe = 5 cm). The ratio f₀/fe is large. In a **microscope**: The objective has a very short focal length (f₀ = 1 cm), and the eyepiece also has a short focal length (fe = 5 cm). Both are short, but the objective is shorter. **Why it matters:** The telescope's long objective focal length allows it to collect light from distant stars and form a small real image very far away. The microscope's very short objective focal length allows it to resolve and magnify tiny nearby specimens. The difference in focal lengths directly determines the working distance (how far the object can be) and the field of view. **(b) Magnification calculations:** **Telescope magnification:** M = f₀/fe = 100/5 = **20×** **Microscope magnification:** M = (f₀ + fe)/(f₀ × fe) × D, or approximately M = D/fe = 25/5 = **5×** (for simple microscope, using magnification ≈ D/fe when object is very close to focal point) **Relationship:** Magnification increases as the objective's focal length increases (for telescopes with fixed eyepiece) and as it decreases (for microscopes). The key insight is that each instrument is optimized for its purpose: telescopes prioritize light collection (large objective), while microscopes prioritize resolution (small objective). **(c) Object placement:** **Telescope:** The object (e.g., a distant star) is at infinity (very far away). The objective forms a real, inverted, diminished image at or near its focal point. **Microscope:** The object (e.g., a bacterial cell) is placed just beyond the focal point of the objective (u ≈ f₀ + small distance). The objective forms a real, inverted, magnified image at a considerable distance inside the microscope tube. **Why positions differ:** The telescope is designed to bring parallel rays (from infinity) to focus, while the microscope is designed to magnify nearby specimens. These opposite optical requirements dictate opposite object placements. **(d) Dim image explanation and solution:** **Why dim:** A very short focal length objective (0.5 cm) has a very small aperture (diameter). Small apertures collect fewer light rays, resulting in a dim image. Additionally, the light is spread over a larger magnified area. **Solution:** - Use a **condenser lens** beneath the specimen to concentrate light onto the object - Use a **brighter light source** (e.g., LED illuminator instead of ambient light) - Increase the **numerical aperture** by using a lens with a larger diameter relative to its focal length - Use **oil immersion** (specialist technique) to increase the light-gathering ability These modifications allow more light to enter the objective and pass through the eyepiece, brightening the final image while preserving magnification.

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