Solved Examples

This section is a dedicated bank of fully worked problems. Formulae you will use again and again:

  • Lens formula: 1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}
  • Power of a lens: P=1fP = \dfrac{1}{f} (with ff in metres, PP in dioptre, D)
  • Myopia: correcting concave lens forms image of infinity at the far point; PP is negative.
  • Hypermetropia: correcting convex lens forms image of the normal near point (25 cm) at the actual near point; PP is positive.

Example 1: Focal length for distant vision (NCERT Q5 i) A person needs a lens of power -5.5 D for distant vision. Find its focal length.

Solution: f=1P=1−5.5=−0.182f = \dfrac{1}{P} = \dfrac{1}{-5.5} = -0.182 m =−18.2= -18.2 cm. It is a concave lens (myopia).

Example 2: Focal length for near vision (NCERT Q5 ii) The same person needs +1.5 D for near vision. Find its focal length.

Solution: f=1P=11.5=0.667f = \dfrac{1}{P} = \dfrac{1}{1.5} = 0.667 m =+66.7= +66.7 cm. It is a convex lens (near-vision/hypermetropia correction).

Example 3: Myopic far point 80 cm (NCERT Q6) The far point of a myopic person is 80 cm in front of the eye. Find the nature and power of the correcting lens.

Solution: Object at infinity, image at far point: v=−80v = -80 cm =−0.8= -0.8 m, u=−∞u = -\infty. 1f=1−0.8−0=−1.25 m−1⇒P=−1.25 D\frac{1}{f} = \frac{1}{-0.8} - 0 = -1.25\ \text{m}^{-1} \Rightarrow P = -1.25\ \text{D} A concave lens of power -1.25 D.

Example 4: Hypermetropic near point 1 m (NCERT Q7) The near point of a hypermetropic eye is 1 m. Find the power of the correcting lens (normal near point 25 cm).

Solution: u=−25u = -25 cm =−0.25= -0.25 m, v=−100v = -100 cm =−1= -1 m. 1f=1−1−1−0.25=−1+4=3 m−1⇒P=+3.0 D\frac{1}{f} = \frac{1}{-1} - \frac{1}{-0.25} = -1 + 4 = 3\ \text{m}^{-1} \Rightarrow P = +3.0\ \text{D} A convex lens of power +3.0 D (f = +0.33 m).

Example 5: Power from focal length A correcting spectacle lens has a focal length of -40 cm. Find its power and nature.

Solution: f=−0.40f = -0.40 m. P=1f=1−0.40=−2.5P = \dfrac{1}{f} = \dfrac{1}{-0.40} = -2.5 D. Concave lens, -2.5 D (corrects myopia).

Example 6: Focal length from power Find the focal length of a lens of power +2.5 D.

Solution: f=1P=12.5=0.40f = \dfrac{1}{P} = \dfrac{1}{2.5} = 0.40 m =+40= +40 cm. A convex lens of focal length 40 cm.

Example 7: Myopic far point 2.5 m A myopic person can see clearly only up to 2.5 m. Find the power of the lens needed.

Solution: v=−2.5v = -2.5 m, u=−∞u = -\infty. 1f=1−2.5=−0.4 m−1⇒P=−0.4\dfrac{1}{f} = \dfrac{1}{-2.5} = -0.4\ \text{m}^{-1} \Rightarrow P = -0.4 D. Concave lens, -0.4 D.

Example 8: Hypermetropic near point 40 cm A hypermetropic eye has near point 40 cm. Find the power of the correcting lens (normal near point 25 cm).

Solution: u=−0.25u = -0.25 m, v=−0.40v = -0.40 m. 1f=1−0.40−1−0.25=−2.5+4=1.5 m−1⇒P=+1.5 D. Convex lens.\frac{1}{f} = \frac{1}{-0.40} - \frac{1}{-0.25} = -2.5 + 4 = 1.5\ \text{m}^{-1} \Rightarrow P = +1.5\ \text{D. Convex lens.}

Example 9: Two lenses in contact Two thin lenses of powers +3 D and -1 D are placed in contact. Find the power and focal length of the combination.

Solution: P=P1+P2=3+(−1)=+2P = P_1 + P_2 = 3 + (-1) = +2 D. f=1P=12=0.5f = \dfrac{1}{P} = \dfrac{1}{2} = 0.5 m =+50= +50 cm. The combination behaves as a convex lens of 50 cm.

Example 10: Combination of a bi-focal-type pair A bi-focal arrangement uses a -2 D and a +5 D lens in contact for a certain test. Find the net power.

Solution: P=−2+5=+3P = -2 + 5 = +3 D. Net focal length f=1/3=0.33f = 1/3 = 0.33 m (convex). (In real bi-focals the two powers act in different regions, but powers of lenses in contact simply add.)

Example 11: Identify the defect (blackboard) A student cannot read the blackboard from the last bench but reads a book easily. Name the defect and the lens.

Solution: Poor distant vision, good near vision → myopia. Corrected by a concave (diverging) lens.

Example 12: Identify the defect (reading) An elderly person holds the newspaper far away to read it. Name the likely defect and the lens.

Solution: Poor near vision → hypermetropia (or presbyopia with age). Corrected by a convex (converging) lens.

Example 13: Range of vision A normal eye has a near point of 25 cm and a far point at infinity. Can it clearly see an object at 20 cm? Explain.

Solution: No. 20 cm is closer than the near point (25 cm). The eye lens cannot reduce its focal length enough to focus it on the retina, so the object looks blurred with strain.

Example 14: Angle of deviation reasoning Why does the emergent ray from a prism not stay parallel to the incident ray, unlike in a glass slab?

Solution: A slab has parallel refracting faces, so bending at the two faces is equal and opposite → emergent ray parallel to incident (only shifted). A prism has inclined faces, so the bendings do not cancel and the emergent ray is deviated by the angle of deviation ∠D\angle D.

Example 15: Which colour deviates more Red and violet light pass through the same prism. Which deviates more and why?

Solution: Violet deviates more. It has the shortest wavelength, for which the prism's refractive index is greater, so it bends the most. Red (longest wavelength) bends the least.

Example 16: Recombining the spectrum How can the VIBGYOR spectrum be recombined into white light?

Solution: By placing a second identical prism inverted (as Newton did) after the first. The second prism deviates the colours in the opposite sense, recombining them so that white light emerges.

Example 17: Rainbow essentials State the position of the Sun and the three processes needed to form a rainbow.

Solution: The Sun must be behind the observer (rainbow appears opposite the Sun). In each water droplet: refraction + dispersion → internal reflection → refraction send the separated colours to the eye.

Example 18: Twinkling vs steady Between a star and a nearby planet, which twinkles and which does not?

Solution: The star twinkles (distant point source; light fluctuates due to atmospheric refraction). The planet does not (nearby extended source; fluctuations from its many points average out).

Example 19: Extra daylight By how long does atmospheric refraction advance sunrise, and what causes it?

Solution: About 2 minutes. The atmosphere bends the light of the Sun (still below the horizon) so that its apparent position is raised above the horizon, letting us see it before it actually crosses the horizon.

Example 20: Sky colour with no air Predict the colour of the sky on the Moon and explain.

Solution: The Moon has no atmosphere, so there is no scattering of sunlight. With no scattered light reaching the eye, the sky on the Moon appears dark (black), even in daytime.

Example 21: Why red for danger Give the optical reason red is chosen for danger and stop signals.

Solution: Red light has the longest wavelength and is scattered the least by fog and smoke. It therefore travels far without much scattering and is seen clearly and in the same colour from a distance.

Example 22: Reddening at sunset Explain why the setting Sun looks red.

Solution: At sunset, sunlight passes through a greater thickness of atmosphere. Most of the shorter (blue) wavelengths are scattered away along the long path, and mainly the longer (red) wavelengths reach the observer, so the Sun appears red.

Example 23: Image distance vs object distance in the eye What happens to the image distance in the eye as an object moves farther away?

Solution: The image is always formed on the retina, so the image distance in the eye stays almost constant. The eye keeps the image on the retina by changing the focal length of the lens (accommodation), not by moving the retina.

Example 24: Power of accommodation defined Define the power of accommodation and name the muscles responsible.

Solution: The power of accommodation is the ability of the eye lens to adjust its focal length to focus objects at different distances on the retina. It is controlled by the ciliary muscles, which change the curvature of the lens.

Example 25: Cataract vs refractive defect How does cataract differ from a refractive defect like myopia, and how is each treated?

Solution: Cataract is a clouding of the crystalline lens that blocks light; it is treated by surgery. Myopia/hypermetropia are refractive defects where the image forms at the wrong place; they are corrected with spectacle lenses (concave/convex).

Example 26: Bi-focal design A person has both myopia and hypermetropia. Describe the spectacle lens they need.

Solution: They need bi-focal lenses: the upper part concave (for distant vision, correcting myopia) and the lower part convex (for near vision, correcting hypermetropia).

Example 27: Myopic far point 100 cm The far point of a myopic eye is 100 cm. Find the power of the correcting lens.

Solution: v=−1v = -1 m, u=−∞u = -\infty. 1f=1−1=−1 m−1⇒P=−1.0\dfrac{1}{f} = \dfrac{1}{-1} = -1\ \text{m}^{-1} \Rightarrow P = -1.0 D. Concave lens, -1.0 D.

Example 28: Hypermetropic near point 75 cm Find the power of a lens to correct a hypermetropic eye whose near point is 75 cm (normal 25 cm).

Solution: u=−0.25u = -0.25 m, v=−0.75v = -0.75 m. 1f=1−0.75−1−0.25=−1.333+4=2.667 m−1⇒P≈+2.67 D. Convex lens.\frac{1}{f} = \frac{1}{-0.75} - \frac{1}{-0.25} = -1.333 + 4 = 2.667\ \text{m}^{-1} \Rightarrow P \approx +2.67\ \text{D. Convex lens.}

Example 29: Least distance of distinct vision What is meant by the least distance of distinct vision, and what is its value for a normal young adult?

Solution: It is the minimum distance at which an object can be seen most distinctly without strain - the near point. For a normal young adult it is about 25 cm.

Example 30: Sky darkness at altitude Why does the sky look darker to a passenger in a high-flying jet than to a person on the ground?

Solution: At high altitude the air is thin, so there are far fewer particles to scatter sunlight. With less scattering, less blue light reaches the eye from the sky, so the sky appears darker than it does at ground level.

Example 31: Combining defect diagnosis and correction A boy can see nearby objects clearly but distant trees look blurred; his far point is 5 m. Diagnose and prescribe.

Solution: He is myopic (far point 5 m, less than infinity). Correcting lens: v=−5v = -5 m, u=−∞u = -\infty → 1f=−0.2 m−1\dfrac{1}{f} = -0.2\ \text{m}^{-1}, so P=−0.2P = -0.2 D, a concave lens of -0.2 D.

Example 32: Why a normal eye can't see below 25 cm Explain why a normal eye cannot clearly see objects placed closer than about 25 cm.

Solution: To focus an object nearer than 25 cm, the eye lens would need a focal length smaller than its minimum limit. The ciliary muscles cannot squeeze the lens that much, so the image no longer forms on the retina and the object appears blurred.