The Human Eye and the Colourful World – Case-based Questions with Answers
20 Case-Based Questions — The Human Eye and the Colourful World
CBSE Board Examination Focus:
- Analyse real-life situations using optics concepts.
- Apply lens formula, dispersion and scattering principles.
- Integrate diagrams, calculations and reasoning in answers.
Instructions: Each case presents a short scenario followed by focused questions and model answers. Use diagrams where indicated and state assumptions clearly when solving numericals.
Case-Based Questions (1–20)
Case 1
Ria notices that while reading a book, the letters become clear only if she holds the book far away (about 60 cm). She complains of blurred near vision but distant objects are clear.
Q: Identify the likely defect and explain why Ria experiences these symptoms.
A: This is hypermetropia (long-sightedness). The eye is unable to focus near objects because the image of near objects would form behind the retina (eye too short or lens too weak). Distant objects are clear because the eye can accommodate for far vision. Correction: convex (converging) lens for near work; power chosen so the near object forms at Ria's near point or at the retina.
A: This is hypermetropia (long-sightedness). The eye is unable to focus near objects because the image of near objects would form behind the retina (eye too short or lens too weak). Distant objects are clear because the eye can accommodate for far vision. Correction: convex (converging) lens for near work; power chosen so the near object forms at Ria's near point or at the retina.
Case 2
A student performing a prism experiment sees the order of colours from red to violet on the screen. However, one of her classmates insists violet should be outermost.
Q: Who is correct and why?
A: The student observing red to violet on the screen is correct: in the primary spectrum produced by a triangular prism, red deviates least and appears on the top/outer side of the spectrum, while violet deviates most and lies on the bottom/inner side. Violet is not outermost because it has the highest refractive index and therefore bends most toward the normal and deviates more.
A: The student observing red to violet on the screen is correct: in the primary spectrum produced by a triangular prism, red deviates least and appears on the top/outer side of the spectrum, while violet deviates most and lies on the bottom/inner side. Violet is not outermost because it has the highest refractive index and therefore bends most toward the normal and deviates more.
Case 3
A person with myopia has a far point of 2 m. He wants spectacles that allow him to see distant objects clearly (objects at infinity).
Q: Calculate the required power of the lens.
A: For object at infinity, image should form at far point v = −2 m (virtual, on same side). Using 1/f = 1/v − 1/u = 1/(−2) − 0 = −1/2 ⇒ f = −2 m. Power P = 1/f = −0.5 D. Therefore a diverging lens of power −0.5 D is prescribed to correct his myopia.
A: For object at infinity, image should form at far point v = −2 m (virtual, on same side). Using 1/f = 1/v − 1/u = 1/(−2) − 0 = −1/2 ⇒ f = −2 m. Power P = 1/f = −0.5 D. Therefore a diverging lens of power −0.5 D is prescribed to correct his myopia.
Case 4
During a hazy afternoon, a photographer notices distant hills appear less sharp and slightly bluish compared to nearer objects.
Q: Explain the optical reasons for haze and bluish tint.
A: The haziness and bluish tint arise due to scattering of light by atmospheric molecules and aerosols. Rayleigh scattering by small molecules preferentially scatters shorter wavelengths (blue) toward the observer, giving distant features a bluish cast. Increased scattering and multiple scattering reduce contrast and sharpness, making distant hills appear hazy.
A: The haziness and bluish tint arise due to scattering of light by atmospheric molecules and aerosols. Rayleigh scattering by small molecules preferentially scatters shorter wavelengths (blue) toward the observer, giving distant features a bluish cast. Increased scattering and multiple scattering reduce contrast and sharpness, making distant hills appear hazy.
Case 5
A teacher asks students to explain why clouds sometimes look white and fluffy even though the sky around is blue.
Q: Provide a concise explanation.
A: Cloud droplets are large compared to wavelength of visible light and scatter all visible wavelengths roughly equally (Mie scattering). The superposition of scattered wavelengths appears white, so clouds look white. The blue sky results from Rayleigh scattering by much smaller air molecules which preferentially scatter blue light.
A: Cloud droplets are large compared to wavelength of visible light and scatter all visible wavelengths roughly equally (Mie scattering). The superposition of scattered wavelengths appears white, so clouds look white. The blue sky results from Rayleigh scattering by much smaller air molecules which preferentially scatter blue light.
Case 6
While doing lens experiments, Meena places an object at 2f from a convex lens and observes the image at 2f on the other side, same size as object.
Q: Explain why the image is same size and give the focal length if object-image distance is 80 cm.
A: For an object at 2f, the image is real, inverted and same size located at 2f. If object-image distance = 80 cm then object at 2f and image at 2f are on opposite sides, so total distance = 4f = 80 cm ⇒ f = 20 cm.
A: For an object at 2f, the image is real, inverted and same size located at 2f. If object-image distance = 80 cm then object at 2f and image at 2f are on opposite sides, so total distance = 4f = 80 cm ⇒ f = 20 cm.
Case 7
A child using a magnifying lens holds it too close to an object and cannot form a clear image on paper.
Q: Why does this happen and how should the child adjust to see a clear magnified virtual image?
A: If the object is within the focal length of the convex lens, the lens produces a virtual, erect, magnified image that cannot be projected on paper. To see a clear magnified virtual image, the child should hold the lens at a distance greater than the focal length (object just inside focal length for larger magnification) and view through the lens with eyes; move lens-object until the virtual image appears sharp to the eye, not on paper.
A: If the object is within the focal length of the convex lens, the lens produces a virtual, erect, magnified image that cannot be projected on paper. To see a clear magnified virtual image, the child should hold the lens at a distance greater than the focal length (object just inside focal length for larger magnification) and view through the lens with eyes; move lens-object until the virtual image appears sharp to the eye, not on paper.
Case 8
During sunset, the Sun appears reddish and the sky near the horizon glows in orange-red hues.
Q: Explain using concepts of scattering why this happens.
A: At sunset sunlight traverses a longer path through the atmosphere; Rayleigh scattering removes shorter wavelengths (blue, green) from the direct path, leaving longer wavelengths (red, orange) to reach the observer. Larger particles and aerosols may enhance Mie scattering, increasing reddening and vivid hues near the horizon.
A: At sunset sunlight traverses a longer path through the atmosphere; Rayleigh scattering removes shorter wavelengths (blue, green) from the direct path, leaving longer wavelengths (red, orange) to reach the observer. Larger particles and aerosols may enhance Mie scattering, increasing reddening and vivid hues near the horizon.
Case 9
A lab group shines white light through a prism and sees a spectrum on the screen. They are asked why they cannot see the same spectrum using a glass slab.
Q: Provide the explanation.
A: A prism has non-parallel faces causing different colours to deviate by different angles — net angular dispersion occurs. A glass slab with parallel faces causes refraction on entry and exit that cancel angular deviation (only lateral shift), so there is negligible angular separation of colours and little or no visible spectrum on a screen.
A: A prism has non-parallel faces causing different colours to deviate by different angles — net angular dispersion occurs. A glass slab with parallel faces causes refraction on entry and exit that cancel angular deviation (only lateral shift), so there is negligible angular separation of colours and little or no visible spectrum on a screen.
Case 10
A person complains of difficulty seeing at night. An eye test reveals reduced rod function.
Q: Explain how rod malfunction leads to night vision problems and suggest possible causes.
A: Rods are highly sensitive photoreceptors essential for vision in dim light; loss of rod function reduces sensitivity to low-intensity light causing night blindness. Causes include vitamin A deficiency, retinitis pigmentosa, retinal degenerative diseases, or trauma. Treatment depends on cause: nutritional correction for vitamin deficiency, medical/clinical intervention otherwise.
A: Rods are highly sensitive photoreceptors essential for vision in dim light; loss of rod function reduces sensitivity to low-intensity light causing night blindness. Causes include vitamin A deficiency, retinitis pigmentosa, retinal degenerative diseases, or trauma. Treatment depends on cause: nutritional correction for vitamin deficiency, medical/clinical intervention otherwise.
Case 11
A student obtains two images by placing an object at different positions relative to a convex lens: one image is real and inverted, another is virtual and erect.
Q: Specify the object positions that produce these images and briefly explain.
A: For a convex lens: if the object is placed beyond the focal length (u > f), the lens forms a real, inverted image on the other side (v positive). If the object is within the focal length (u < f), the lens produces a virtual, erect, magnified image on the same side as the object (v negative when using sign convention). Thus object beyond f → real inverted; object within f → virtual erect.
A: For a convex lens: if the object is placed beyond the focal length (u > f), the lens forms a real, inverted image on the other side (v positive). If the object is within the focal length (u < f), the lens produces a virtual, erect, magnified image on the same side as the object (v negative when using sign convention). Thus object beyond f → real inverted; object within f → virtual erect.
Case 12
A person standing near a fountain sees a rainbow on a sunny day. The observer notes the angle of the bow relative to the antisolar point.
Q: Explain the geometry and typical angle for a primary rainbow and why the colours are ordered with red outside.
A: A primary rainbow is formed by one internal reflection inside raindrops; different colours emerge at slightly different angles — red at about 42° to the observer's line from the antisolar point (centre of the circular bow), violet at about 40°. Red appears on the outer edge because it deviates the least. The circular geometry arises because only droplets at specific angles relative to the observer reflect light into the eye, producing a cone; intersection with ground/observer plane gives an arc.
A: A primary rainbow is formed by one internal reflection inside raindrops; different colours emerge at slightly different angles — red at about 42° to the observer's line from the antisolar point (centre of the circular bow), violet at about 40°. Red appears on the outer edge because it deviates the least. The circular geometry arises because only droplets at specific angles relative to the observer reflect light into the eye, producing a cone; intersection with ground/observer plane gives an arc.
Case 13
A student records that a spectacle lens has a power +2.5 D. He is asked what is its focal length and whether it's converging or diverging.
Q: Compute focal length and identify lens type.
A: Power P = +2.5 D ⇒ f = 1/P = 1/2.5 = 0.4 m = 40 cm. Positive power indicates a converging (convex) lens.
A: Power P = +2.5 D ⇒ f = 1/P = 1/2.5 = 0.4 m = 40 cm. Positive power indicates a converging (convex) lens.
Case 14
During an optics practical, students must explain why sky colour depends on particle size and concentration in the atmosphere.
Q: Give an answer relating Rayleigh and Mie scattering.
A: Rayleigh scattering predominates for particles much smaller than wavelength (air molecules) and is strongly wavelength-dependent (∝1/λ^4), producing blue skies. Mie scattering occurs for larger particles (aerosols, dust) and is less wavelength-selective, scattering all colours similarly and causing pale/white/gray skies and haze. Particle concentration influences scattering intensity and visibility; more particles increase scattering and can change sky colour and contrast.
A: Rayleigh scattering predominates for particles much smaller than wavelength (air molecules) and is strongly wavelength-dependent (∝1/λ^4), producing blue skies. Mie scattering occurs for larger particles (aerosols, dust) and is less wavelength-selective, scattering all colours similarly and causing pale/white/gray skies and haze. Particle concentration influences scattering intensity and visibility; more particles increase scattering and can change sky colour and contrast.
Case 15
A student is correcting myopia with spectacles and asks why lenses look thicker at the center for some prescriptions and thinner for others.
Q: Explain thickness variation with lens type.
A: Converging (convex) lenses have greater curvature at centre and thus are thicker at the centre than at edges; used for hypermetropia. Diverging (concave) lenses are thinner at centre and thicker at edges; used for myopia. Thickness depends on required power (higher power → higher curvature → more thickness), material refractive index (higher index allows thinner lenses), and cosmetic design.
A: Converging (convex) lenses have greater curvature at centre and thus are thicker at the centre than at edges; used for hypermetropia. Diverging (concave) lenses are thinner at centre and thicker at edges; used for myopia. Thickness depends on required power (higher power → higher curvature → more thickness), material refractive index (higher index allows thinner lenses), and cosmetic design.
Case 16
A child looks at a red toy under a green lamp and says the toy looks black.
Q: Explain optical reasoning.
A: The red toy reflects red wavelengths and absorbs others. A green lamp emits mostly green wavelengths; since little red light is present to be reflected, the red toy reflects little light and appears dark/black under green illumination. This demonstrates that observed colour depends on both object reflectance and incident light spectrum.
A: The red toy reflects red wavelengths and absorbs others. A green lamp emits mostly green wavelengths; since little red light is present to be reflected, the red toy reflects little light and appears dark/black under green illumination. This demonstrates that observed colour depends on both object reflectance and incident light spectrum.
Case 17
A student observes that the spectrum from a prism is fainter after passing through a roughened glass surface prior to the prism.
Q: Why does roughening reduce brightness of spectrum?
A: Roughening scatters incoming light diffusely, reducing the intensity of the collimated beam entering the prism. Since dispersion requires a strong incident beam to produce a bright spectrum, the scattered, less-collimated light yields a fainter, less distinct spectrum on the screen. Surface roughness increases losses due to scattering and reduces coherent energy directed through prism faces.
A: Roughening scatters incoming light diffusely, reducing the intensity of the collimated beam entering the prism. Since dispersion requires a strong incident beam to produce a bright spectrum, the scattered, less-collimated light yields a fainter, less distinct spectrum on the screen. Surface roughness increases losses due to scattering and reduces coherent energy directed through prism faces.
Case 18
An optometrist measures a patient's near point as 50 cm. The patient wants to use spectacles for comfortable reading at 25 cm.
Q: Calculate the power of spectacles required for near work.
A: We need a lens that makes an object at 25 cm appear at 50 cm (patient's near point). Using lens formula with u = −0.25 m, v = −0.50 m (virtual image at near point): 1/f = 1/v − 1/u = 1/(−0.50) − 1/(−0.25) = (−2) + 4 = 2 ⇒ f = 0.5 m. Power P = 1/f = +2.0 D. So spectacles of +2.0 D are required for comfortable reading at 25 cm.
A: We need a lens that makes an object at 25 cm appear at 50 cm (patient's near point). Using lens formula with u = −0.25 m, v = −0.50 m (virtual image at near point): 1/f = 1/v − 1/u = 1/(−0.50) − 1/(−0.25) = (−2) + 4 = 2 ⇒ f = 0.5 m. Power P = 1/f = +2.0 D. So spectacles of +2.0 D are required for comfortable reading at 25 cm.
Case 19
While working with ray diagrams, a student forgets to mark the direction arrows on rays and submits the diagram in an exam.
Q: Discuss the importance of direction arrows and likely impact on evaluation.
A: Direction arrows show the path of light (incident vs emergent), clarify whether rays are real or extensions (use dotted lines), and avoid ambiguity in ray tracing. Omitting arrows can confuse the examiner about which side rays travel and may lead to loss of method marks. Always mark arrows, use dotted lines for virtual rays and label focal points to maximise marks.
A: Direction arrows show the path of light (incident vs emergent), clarify whether rays are real or extensions (use dotted lines), and avoid ambiguity in ray tracing. Omitting arrows can confuse the examiner about which side rays travel and may lead to loss of method marks. Always mark arrows, use dotted lines for virtual rays and label focal points to maximise marks.
Case 20
A group models rainbow formation and a student asks why a secondary rainbow is fainter and has reversed colours compared to the primary.
Q: Provide a clear explanation.
A: A secondary rainbow is formed by two internal reflections inside raindrops. Each reflection reduces intensity (some light exits at each interface), so the secondary bow is fainter. The second reflection reverses the order of colours: red appears on the inner edge of the secondary bow because the sequence of deviations causes the angular ordering to invert. The secondary rainbow appears at a larger angle (about 51°) and is less bright and more diffuse.
A: A secondary rainbow is formed by two internal reflections inside raindrops. Each reflection reduces intensity (some light exits at each interface), so the secondary bow is fainter. The second reflection reverses the order of colours: red appears on the inner edge of the secondary bow because the sequence of deviations causes the angular ordering to invert. The secondary rainbow appears at a larger angle (about 51°) and is less bright and more diffuse.