Light – Reflection and Refraction – Case-based Questions with Answers
CBSE Class 10 Physics — Chapter 9: Light — 20 Case-Based Questions & Answers
Class: CBSE Class 10
Subject: Physics
Chapter: Chapter 9 — Light: Reflection & Refraction
CBSE Board Examinations
Systematic order: Concepts → Ray diagrams → Formulas → Applications → Numerical reasoning
Instructions: 20 case-based questions (scenario + focused questions) with clear, exam-oriented answers. All content is aligned with NCERT Class 10 Physics Chapter 9. Use these for classroom discussion, practice and board exam revision.
Case 1 — Mirror in a dressing room: A student notices the image in the dressing mirror looks the same size but appears behind the mirror at equal distance.
Q1.1: Which type of mirror produces such an image and what are its characteristics?
A: A plane mirror produces a virtual, erect image of the same size as the object. Image distance equals object distance and the image is laterally inverted.
Q1.2: Explain lateral inversion and give one practical consequence.
A: Lateral inversion is the left–right reversal seen in plane mirror images. Practical consequence: text appears reversed, so mirror-reading requires reversing letters or using mirrored text.
Case 2 — Car side mirror: A motorist complains that objects appear smaller but more of the road is visible in the side mirror.
Q2.1: Which mirror is typically used and why?
A: A convex mirror is used because it produces a virtual, erect and diminished image giving a wider field of view — useful for observing a larger area behind the vehicle.
Q2.2: Explain how curvature affects field of view and image size.
A: A mirror with smaller radius (more curved) produces greater divergence of reflected rays, producing smaller images but a larger field of view. Convex curvature reduces image size and increases coverage area.
Case 3 — Science lab experiment: Students focus sunlight with a concave mirror and notice a hot spot at a definite point.
Q3.1: Name the point where parallel rays converge and give its relation to radius of curvature R.
A: The point is the focal point F. For a spherical concave mirror focal length f = R/2, where R is the radius of curvature.
Q3.2: Why does a concave mirror concentrate sunlight and how is it used practically?
A: Concave mirror reflects parallel rays to the focal point (convergence), concentrating energy and producing heat — used in solar cookers, searchlights and concentrating collectors.
Case 4 — A camera lens forms a sharp image on film when focused at various distances.
Q4.1: Which lens property allows changing focus for different object distances?
A: The lens formula 1/v − 1/u = 1/f (or 1/v + 1/u = 1/f depending on sign convention) relates object distance u, image distance v and focal length f. Adjusting distance between lens and film (v) focuses images from different u.
Q4.2: If a lens has f = 50 mm and an object is far away, where should the image be formed?
A: For a distant object (u → ∞), image forms at the focal plane, v ≈ f (≈ 50 mm). This is why camera sensors/film are placed at the focal length for infinity focus.
Case 5 — Pool distortions: A person notices a coin at the bottom of a pool appears closer than it really is.
Q5.1: Explain the phenomenon and provide the relation involving refractive index for near-normal viewing.
A: Apparent depth due to refraction: objects in water appear raised because light from the object refracts at the water–air interface and emerges toward the eye. For near-normal incidence, refractive index n ≈ real depth / apparent depth.
Q5.2: If real depth is 2.0 m and n (water) = 1.33, what is approximate apparent depth?
A: Apparent depth ≈ real depth / n = 2.0 / 1.33 ≈ 1.50 m. The coin appears about 0.5 m higher than its real position.
Case 6 — Bent straw: A straw in a glass appears bent at the surface of water.
Q6.1: Which optical process explains this and why does the straw look broken?
A: Refraction causes bending of light at the water–air interface because speed in water differs from air; the portion of straw underwater appears displaced due to change in direction of light rays, giving a 'broken' appearance.
Q6.2: Would the effect change if the glass were filled with oil of refractive index closer to that of air?
A: Yes — if refractive index difference is smaller, the amount of bending reduces and the apparent break decreases. For n closer to air (≈1), refraction is less noticeable.
Case 7 — Optical fibre endoscopy: A hospital uses fibre optic cables to view internal organs.
Q7.1: Explain how optical fibres transmit light over curved paths inside the body.
A: Optical fibres use total internal reflection (TIR). Core has higher refractive index than cladding; light entering within the acceptance angle reflects repeatedly at core–cladding boundary with incidence > critical angle, enabling transmission around bends with low loss.
Q7.2: State two design requirements for medical fibres.
A: (1) Flexible, low-attenuation core and protective cladding; (2) Biocompatible, sterilizable outer jackets and coherent bundles for image transmission (ordered cores to preserve spatial information).
Case 8 — TIR in diamond: A jeweller explains why diamonds sparkle intensely.
Q8.1: How does refractive index and TIR contribute to diamond brilliance?
A: Diamond has very high refractive index (≈ 2.42) producing a small critical angle; internal reflections (TIR) trap light, causing multiple internal bounces and strong dispersion — resulting in sparkle and brilliance.
Q8.2: Would a material with low refractive index show similar sparkle? Why/why not?
A: No — low refractive index gives larger critical angle, reducing internal reflection; less light is trapped, so sparkle diminishes.
Case 9 — Laser alignment: A laser beam through a glass slab emerges parallel but displaced.
Q9.1: Name and explain the effect observed and why emergent ray is parallel to incident ray.
A: Lateral displacement in a parallel-sided slab: the ray refracts on entering and exiting; because slab faces are parallel, emergent ray is parallel to incident ray but shifted sideways. Displacement depends on thickness and angles via geometric relations.
Q9.2: How can lateral displacement be minimised in precision optical setups?
A: Use thinner slabs, set near-normal incidence (small angles), or use anti-reflection coatings and precision-mounted components to reduce unwanted offsets.
Case 10 — Lens magnifier: A student uses a convex lens as a magnifier to read small print.
Q10.1: What must be the object position relative to focal length to get a virtual magnified image?
A: Place the object within the focal length (u < f) of the convex lens. The lens produces a virtual, erect and magnified image on the same side as the object.
Q10.2: Why does the eye perceive a larger angular size with the magnifier?
A: The magnifier increases the angular size of the object subtended at the eye by producing a virtual image closer to the eye or of larger size, making details resolvable with less accommodation.
Case 11 — Classroom ray diagram: A student draws only one ray and gets partial marks.
Q11.1: Which principal rays should be drawn for full credit when locating an image for lens/mirror?
A: Draw at least two principal rays: (1) ray parallel to principal axis (then through focus), and (2) ray through optical centre (undeviated) for lenses; for mirrors, use parallel → focus and through centre → back. Two rays are sufficient to locate image intersection.
Q11.2: Give one exam tip for neat ray diagrams.
A: Use thin lines, label P, F, O, C and arrowheads, indicate object and image heights and state image nature (real/virtual, erect/inverted) clearly beneath the diagram.
Case 12 — Aberration problem: A telescope shows blurred images at edges when using a simple spherical mirror.
Q12.1: Identify the optical aberration and why it occurs.
A: Spherical aberration — spherical surfaces do not focus paraxial and marginal rays at same point, causing a blurred focal region. It occurs because mirror shape deviates from ideal parabolic geometry for off-axis rays.
Q12.2: Suggest two remedies used in optical instruments.
A: Use parabolic mirrors, aspheric surfaces, or stop down aperture (use only central rays); multi-element corrective lens systems also reduce aberration.
Case 13 — Aquarium viewing: Observers at different angles see fish at different apparent positions when looking through glass.
Q13.1: Explain dependence of apparent position on viewing angle using refraction concepts.
A: Apparent position depends on refraction angle: Snell’s law (n₁ sin i = n₂ sin r) dictates bending amount. For oblique viewing, larger i causes more deviation, changing the emergent ray direction and hence perceived position; apparent depth varies with angle.
Q13.2: How does anti-reflective coating or using wider viewing windows help observers?
A: Wider viewing areas reduce strong edge-angle refraction and coatings reduce surface reflections improving clarity; minimal glass thickness and placing observer near normal incidence reduces apparent distortion.
Case 14 — Safety: A school warning not to focus sunlight onto skin using magnifying lens.
Q14.1: Explain the hazard using optics principles.
A: A convex lens converges parallel sunlight to the focal point concentrating energy and raising local intensity and temperature — causing burns or fire. Focal spot heating is hazardous.
Q14.2: Suggest two safety precautions for optics practicals involving concentrated light.
A: Never focus sunlight on flammable materials or skin; supervise experiments, use small test areas with heat-resistant mounts and avoid leaving focused beams unattended. Use low-power sources when possible.
Case 15 — Measuring focal length: Students use distant object method but get slightly different f each time.
Q15.1: List three experimental errors that can affect focal length measurement using sunlight.
A: (1) Parallax while measuring lens–screen distance; (2) imperfect alignment of lens normal to incoming rays; (3) spherical aberration or using non-paraxial rays (large aperture) leading to indistinct focal spot.
Q15.2: How can students improve accuracy?
A: Use a lens holder for alignment, use an aperture to restrict to paraxial rays, take multiple measurements and average, and measure distance from optical centre rather than lens edge.
Case 16 — Refractive index check: A lab reports n for a glass slab using measured angles; one reading seems off.
Q16.1: Which systematic errors could cause erroneous refractive index from Snell’s law?
A: Misalignment of protractor/incident ray, incorrect normal construction, inaccurate angle measurement due to parallax, or using a non-uniform slab causing internal scattering.
Q16.2: Suggest one quick check to validate measurements.
A: Repeat at multiple incident angles, plot sin i vs sin r — slope should be close to constant; outlier points indicate measurement error to be rechecked.
Case 17 — Rainbow & scattering: A student asks why we see a rainbow but not a single refracted beam of all colours together.
Q17.1: Explain dispersion and how it leads to rainbow formation.
A: Dispersion: refractive index depends on wavelength; different colours refract by different amounts. In raindrops sunlight is refracted, reflected internally and refracted again, separating colours spatially to form a spectrum (rainbow) with violet deviated most and red least.
Q17.2: Why does white beam separate but sunlight still looks white in ordinary refraction experiments?
A: In ordinary refraction, dispersion is small or overlapping; also the observer may combine all refracted colours unless setup separates them spatially (prism or raindrop geometry) producing visible spectrum.\
Case 18 — Vehicle headlight alignment: A mechanic adjusts headlights to avoid glare to oncoming drivers.
Q18.1: Which optical principle guides correct alignment of headlights?
A: Headlights should be aligned so that beams are parallel (or aimed slightly downwards) — using reflectors and parabolic surfaces ensures parallel output; correct aim prevents dazzling oncoming drivers by controlling beam direction.
Q18.2: How does beam divergence affect road illumination and glare?
A: Narrow, well-collimated beams illuminate distant road but may glare; wider beams illuminate near field. Proper reflector/lens design balances divergence to provide safe illumination without causing glare.
Case 19 — Rear-projection TV: A screen uses lenses to project enlarged images from a small source.
Q19.1: Which lens arrangement (convex/concave) is used for projection and why?
A: A projection system uses convex (converging) lenses to form a real, enlarged image on a screen. Large image size achieved by placing object (microdisplay) between f and 2f to produce a magnified real image.
Q19.2: Mention two optical challenges in projection systems.
A: (1) Aberrations (spherical, chromatic) requiring corrective optics; (2) brightness loss requiring high-intensity illumination and efficient lens coatings to reduce reflection losses.
Case 20 — Exam question practice: A teacher asks why image nature (real/virtual, erect/inverted) must be stated clearly in answers.
Q20.1: Why is specifying image nature important in optics answers?
A: Nature indicates how the image is formed (by actual convergence or apparent divergence), whether it can be projected (real) or only seen (virtual), and its orientation (erect/inverted) — key to full conceptual understanding and marks in board answers.
Q20.2: Give a short checklist students should follow when solving mirror/lens problems in exams.
A: Checklist: (1) State formula and sign convention; (2) Substitute values and show algebra; (3) Compute magnification and interpret sign; (4) State image nature and draw a labelled ray diagram; (5) Mention assumptions (thin lens, paraxial approximation).
Tip: For case-based answers in exams, identify the core optical principle (reflection, refraction, TIR, lens/mirror formula), explain cause–effect briefly, show one calculation or diagram if relevant, and conclude with the practical implication.