Question 1 · MCQ · Analyse · 1 mark
Distant writing looks blurred
Defects of vision
A student sees a nearby book clearly but distant writing is blurred. Which defect and correcting lens are most consistent with this description?
(A) Hypermetropia; convex lens
(B) Myopia; concave lens
(C) Myopia; convex lens
(D) Hypermetropia; concave lens
Show answer and solution
Answer: (B) Myopia; concave lens
Worked solution: In myopia, distant objects tend to focus in front of the retina. A concave lens diverges the entering rays so the eye can focus them on the retina.
Question 2 · MCQ · Analyse · 1 mark
White light through a prism
Dispersion of light
A narrow beam of white light passes through a glass prism and forms a spectrum. Which colour usually deviates the most?
(A) Red
(B) Yellow
(C) Violet
(D) Green
Show answer and solution
Answer: (C) Violet
Worked solution: For ordinary glass in the visible range, violet light has the greatest refractive index and deviates more than red light.
Question 3 · MCQ · Analyse · 1 mark
Looking from far to near
Human eye and accommodation
A normal eye changes focus from a distant tree to a nearby book. What usually happens to the eye lens?
(A) It becomes more curved and its focal length decreases.
(B) It becomes less curved and its focal length increases.
(C) It leaves the eyeball.
(D) Its focal length remains fixed in all cases.
Show answer and solution
Answer: (A) It becomes more curved and its focal length decreases.
Worked solution: Ciliary muscles alter lens curvature for accommodation. For a near object, the lens becomes thicker or more curved, increasing power and reducing focal length.
Question 4 · MCQ · Analyse · 1 mark
Why a warning light is red
Scattering of light
One reason red is suitable for a distant warning light is that, compared with blue light, it is generally:
(A) Scattered more strongly by very small particles.
(B) Unable to travel through air.
(C) Always brighter regardless of lamp power.
(D) Scattered less strongly by very small particles.
Show answer and solution
Answer: (D) Scattered less strongly by very small particles.
Worked solution: Red has a longer wavelength and is scattered less strongly by particles much smaller than visible wavelengths. This helps it remain visible over distance; actual visibility also depends on conditions and lamp design.
Question 5 · A-R · Analyse · 1 mark
A blue daytime sky
Scattering of light
Assertion (A): The daytime sky often appears blue away from the Sun.
Reason (R): Air molecules scatter shorter visible wavelengths more strongly than longer visible wavelengths.
(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not correctly explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show answer and solution
Answer: (A) Both A and R are true, and R correctly explains A.
Worked solution: Scattered short-wavelength light reaches the eye from many directions. The observed colour depends on the sunlight spectrum and human vision as well, so the sky generally looks blue rather than violet.
Question 6 · A-R · Analyse · 1 mark
Retina and lens function
Human eye and accommodation
Assertion (A): The retina changes the curvature of the eye lens during accommodation.
Reason (R): Ciliary muscles help change the curvature of the eye lens.
(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not correctly explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show answer and solution
Answer: (D) A is false, but R is true.
Worked solution: The retina detects light. Ciliary muscles change lens shape to adjust focus, so the reason is true and the assertion is false.
Question 7 · VSA · Apply · 2 marks
Reading a negative prescription
Defects of vision
A correcting lens is marked −2.0 D. State the lens type and calculate its focal length in metres.
Show answer and solution
Answer: Concave lens; f = −0.50 m.
Worked solution: A negative power identifies a diverging lens. Using f = 1/P gives f = 1/(−2.0) = −0.50 m.
Question 8 · VSA · Apply · 2 marks
Recombining a spectrum
Dispersion of light
A second identical prism is placed inverted after the first prism so that it receives all the separated colours. What can be obtained and what does this show?
Show answer and solution
Answer: Approximately white light can be recombined, showing that white light contains different colours.
Worked solution: The inverted prism counteracts the first prism’s separation under suitable alignment. The colours recombine instead of being newly created by the prism.
Question 9 · SA · Analyse · 3 marks
Correcting a finite far point
Defects of vision
A myopic person has a far point 2 m from the eye. Neglect the distance between spectacles and eye. Calculate the spectacle-lens power needed to see a very distant object.
Show answer and solution
Answer: −0.50 D.
Worked solution: A distant object must have a virtual image at the person’s far point, so v = −2 m and u is effectively −∞. The lens formula gives f = −2 m. Thus P = 1/f = −0.50 D, requiring a concave lens.
Question 10 · SA · Analyse · 3 marks
Correcting a distant near point
Defects of vision
A person’s near point is 50 cm. What lens power allows reading at 25 cm, neglecting the distance between spectacles and eye?
Show answer and solution
Answer: +2.0 D, using a convex lens.
Worked solution: The lens must form a virtual image at v = −50 cm for an object at u = −25 cm. From 1/f = 1/v − 1/u, 1/f = −1/50 + 1/25 = 1/50 cm⁻¹. Hence f = +50 cm = +0.50 m and P = +2.0 D.
Question 11 · LA · Analyse · 5 marks
Distinguishing two vision defects
Human eye and accommodation
For myopia and hypermetropia, compare (i) the distance range that is difficult to see clearly, (ii) the image position relative to the retina without correction, (iii) one possible structural or optical cause, (iv) correcting lens type, and (v) how that lens changes the incident rays.
Show answer and solution
Answer: Myopia: distant objects, focus in front, excessive eye power or elongated eyeball, concave correction. Hypermetropia: near objects, focus behind, insufficient power or shortened eyeball, convex correction.
Worked solution: A myopic eye may have excessive converging power or an elongated eyeball. A concave lens diverges rays before they enter, moving the final focus back onto the retina. A hypermetropic eye may have insufficient converging power or a shorter eyeball. A convex lens pre-converges rays from a nearby object so the eye can focus them on the retina.
Question 12 · Case Study · Analyse · 4 marks
Comparing two spectacle lenses
Defects of vision
Lens P is labelled −1.5 D and lens Q is labelled +2.5 D. Consider the usual school-level corrections for myopia and hypermetropia.
Answer the following:
(i) Which lens is diverging? [1 mark]
(ii) Which lens is suitable for correcting hypermetropia in this model? [1 mark]
(iii) Find both focal lengths in metres and interpret their signs. [2 marks]
Show answer and solution
Answer: (i) P. (ii) Q. (iii) fP ≈ −0.67 m and fQ = +0.40 m; negative is diverging and positive is converging.
Worked solution: Using f = 1/P gives −1/1.5 ≈ −0.67 m for P and 1/2.5 = +0.40 m for Q. A concave lens corrects the excess effective convergence of a myopic eye; a convex lens helps a hypermetropic eye focus near objects.
