Light — reflection, refraction, dispersion, human eye defects

Question

A student cannot see distant objects clearly but can read a book held close. (a) What defect of vision does the student have? (b) What type of lens is used to correct it? (c) If the far point is 2 m, what should be the power of the corrective lens?

(CBSE Class 10 — Human Eye and the Colourful World)

Light Phenomenon Decision Tree

flowchart TD
    A["Light Problem"] --> B{Which phenomenon?}
    B -->|Bouncing off surface| C["Reflection"]
    B -->|Bending at boundary| D["Refraction"]
    B -->|Splitting into colours| E["Dispersion"]
    B -->|Eye cannot see clearly| F["Vision Defect"]
    C --> C1["Laws: angle i = angle r"]
    D --> D1["Snell's Law: n1 sin i = n2 sin r"]
    E --> E1["Prism separates white light"]
    F --> F1{"Cannot see far?"}
    F --> F2{"Cannot see near?"}
    F1 -->|Yes| G["Myopia — use concave lens"]
    F2 -->|Yes| H["Hypermetropia — use convex lens"]

Solution — Step by Step

The student cannot see distant objects but can see nearby objects clearly. This means the image of distant objects forms before the retina (not on it). This defect is called myopia (short-sightedness or near-sightedness).

In myopia, the eyeball is too long or the eye lens is too curved, so the focal length is too short.

We need a lens that diverges the incoming light slightly, so the image shifts back onto the retina. A concave lens (diverging lens) does this.

The concave lens takes parallel rays from a distant object and makes them appear to come from the far point of the eye.

The far point is 2 m. The concave lens must form a virtual image at 2 m for an object at infinity.

Using the lens formula with $u = \infty$ (object at infinity) and $v = -2$ m (image at far point, same side as object for concave lens):

\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-2} - \frac{1}{\infty} = -\frac{1}{2}

f = -2 \text{ m}

P = \frac{1}{f} = \frac{1}{-2} = \mathbf{-0.5 \text{ D (dioptre)}}

\boxed{P = -0.5 \text{ D}}

The negative sign confirms it is a concave (diverging) lens.

Why This Works

A normal eye adjusts its lens shape (accommodation) to focus objects at different distances — from 25 cm (near point) to infinity (far point). In myopia, the far point moves closer than infinity. The concave lens compensates by making distant objects appear to be at the far point, so the eye can focus on them.

The power $P = 1/f$ measures how strongly the lens bends light. Negative power means diverging. The larger the magnitude, the stronger the correction needed.

Alternative Method — Direct Formula

For myopia correction, the focal length of the lens equals the far point distance (with a negative sign):

f = -d_{\text{far point}} = -2 \text{ m}

P = -0.5 \text{ D}

This shortcut works because the lens must map infinity to the far point.

For CBSE Class 10, remember: Myopia = concave lens (negative power). Hypermetropia = convex lens (positive power). Presbyopia (age-related) = bifocal lens. These three defects with their corrections are guaranteed questions worth 3-5 marks every year.

Common Mistake

Students often confuse myopia and hypermetropia. Quick check: if the person cannot see far objects, the defect starts with a different letter than “far” — it is myopia (M for myopia, not far). If the person cannot see near objects, it is hypermetropia (H). Also, do not forget the negative sign for the concave lens focal length — writing $P = +0.5$ D instead of $-0.5$ D changes the lens type entirely.