Human Eye & Colourful World (Numerical with Answer)

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Home CBSE Class 10 Science Important Questions for Class 10 Science Human Eye & Colourful World

Here we are providing numerical assignment on class 10 science chapter human eye and colourful world. Answers are also provided for reference.

Table of Contents

Numerical Problems on Human Eye and Colourful World

Class 10 Science · The Human Eye

Eye Defects Practice Set 1

Apply lens formula $\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$ for vision correction

Ch 11 · Human Eye

If the concave lens of focal length (f = 1.5m) used to restore the proper vision, then what is the power of lens?

Answer & Solution

Answer: $-0.67\text{ D}$

Explanation: A concave lens always has a negative focal length. So, $f = -1.5\text{ m}$.

The power of a lens is the reciprocal of its focal length in meters:

$$P = \frac{1}{f} = \frac{1}{-1.5}$$

$$P = -\frac{10}{15} = -\frac{2}{3} \approx -0.67\text{ D}$$

A young boy can adjust the power of his eye-lens between 50 D & 60 D. His far point is infinity.
(a) What is the distance of his retina from the eye-lens?
(b) What is his near point?

Answer & Solution

Answer: (a) $2\text{ cm}$ (b) $10\text{ cm}$

Explanation:
(a) When the eye is fully relaxed (looking at infinity), its focal length is largest, meaning the power is at its minimum (50 D). The parallel rays from infinity focus exactly on the retina.
Therefore, the distance to the retina is equal to this maximum focal length:

$$f_{max} = \frac{1}{P_{min}} = \frac{1}{50}\text{ m} = 0.02\text{ m} = 2\text{ cm}$$

(b) When looking at the near point, the eye uses its maximum power (60 D). The image distance $v$ remains fixed at the retina distance ($v = +2\text{ cm} = +0.02\text{ m}$).

$$P_{max} = \frac{1}{v} – \frac{1}{u} \implies 60 = \frac{1}{0.02} – \frac{1}{u}$$

$$60 = 50 – \frac{1}{u} \implies \frac{1}{u} = 50 – 60 = -10$$

$$u = -0.1\text{ m} = -10\text{ cm}$$

His near point is at $10\text{ cm}$ in front of the eye.

A person cannot see objects clearly beyond 50 cm. Find the power of the lens to correct the vision.

Answer & Solution

Answer: $-2\text{ D}$

Explanation: The person is myopic (short-sighted). The corrective lens must take an object from infinity ($u = -\infty$) and form a virtual image at the person’s far point ($v = -50\text{ cm}$).

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-50} – \frac{1}{-\infty} = -\frac{1}{50} – 0$$

So, $f = -50\text{ cm} = -0.5\text{ m}$.

$$P = \frac{1}{f} = \frac{1}{-0.5} = -2\text{ D}$$

A myopic person having far point 80 cm uses spectacles of power –1.0 D. How far can he see clearly?

Answer & Solution

Answer: $400\text{ cm}$ (or $4\text{ m}$)

Explanation: The focal length of the prescribed spectacles is $f = \frac{1}{P} = \frac{1}{-1.0} = -1\text{ m} = -100\text{ cm}$.
With these spectacles, the virtual image of the furthest object he can see must form at his actual far point ($v = -80\text{ cm}$). We need to find the new maximum object distance ($u$).

$$\frac{1}{u} = \frac{1}{v} – \frac{1}{f} = \frac{1}{-80} – \frac{1}{-100}$$

$$\frac{1}{u} = -\frac{5}{400} + \frac{4}{400} = -\frac{1}{400}$$

Therefore, $u = -400\text{ cm}$. He can see clearly up to a distance of $4\text{ m}$.

A person having a myopic eye used the concave lens of focal length 50 cm. What is the power of the lens?

Answer & Solution

Answer: $-2.0\text{ D}$

Explanation: A concave lens has a negative focal length. $f = -50\text{ cm} = -0.5\text{ m}$.

$$P = \frac{1}{f} = \frac{1}{-0.5} = -2.0\text{ D}$$

Learn Human Eye and Colourful World through Animations

Eye Defects Practice Set 2

Apply lens formula $\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$ for vision correction

Ch 11 · Human Eye

A 52-year-old near-sighted person wears eye glass of power of –5.5D for distance viewing. His doctor prescribes a correction of +1.5D in the near-vision section of his bi-focals this measured relative to the main parts of the lens.
(i) What is the focal length of his distance viewing part of the lens?
(ii) What is the focal length of the near vision section of the lens?

Answer & Solution

Answer: (i) $-18.2\text{ cm}$ (ii) $-25\text{ cm}$

Explanation:
(i) Distance viewing: The power of the main (upper) part of the lens is $P_1 = -5.5\text{ D}$.

$$f_1 = \frac{1}{P_1} = \frac{1}{-5.5} \approx -0.1818\text{ m} = -18.2\text{ cm}$$

(ii) Near vision: The addition (ADD) for near vision is measured relative to the main distance prescription. Therefore, the total power of the lower near-vision section is $P_{near} = P_{distance} + P_{add}$.

$$P_2 = -5.5\text{ D} + (+1.5\text{ D}) = -4.0\text{ D}$$

$$f_2 = \frac{1}{P_2} = \frac{1}{-4.0} = -0.25\text{ m} = -25\text{ cm}$$

A short-sighted person cannot see clearly beyond 2 m. Calculate the power of lens required to correct his vision.

Answer & Solution

Answer: $-0.5\text{ D}$

Explanation: The person is short-sighted (myopic) with a far point at $v = -2\text{ m}$. To correct this, the lens must form a virtual image at $2\text{ m}$ for an object placed at infinity ($u = -\infty$).

$$\frac{1}{f} = \frac{1}{-2} – \frac{1}{-\infty} \implies f = -2\text{ m}$$

$$P = \frac{1}{f} = \frac{1}{-2} = -0.5\text{ D}$$

A person is able to see objects clearly only when these are lying at distance between 50 cm and 300 cm from his eye.
(i) What kind of defect of vision he is suffering from?
(ii) What kind of lenses will be required to increase his range of vision from 25 cm to infinity? Explain briefly.

Answer & Solution

Explanation:
(i) For a normal eye, the near point is at $25\text{ cm}$ and the far point is at infinity. Since this person cannot see objects clearly if they are closer than $50\text{ cm}$ or further than $300\text{ cm}$, he is suffering from both Myopia (short-sightedness) and Hypermetropia (long-sightedness). This dual condition is often associated with Presbyopia in aging eyes.

(ii) A bi-focal lens consisting of both a concave and convex lens will be required. The upper portion of the spectacle will contain a concave lens of suitable focal length to push the far point to infinity (correcting myopia). The lower portion will contain a convex lens to bring the near point back to $25\text{ cm}$ (correcting hypermetropia).

A 14-years old student is not able to see clearly to question written on a black board placed at a distance of 5 m from him.
(a) Name the defect of vision he is suffering from.
(b) Name the type of lens used to correct this defect.
(c) State two causes of this defect.

Answer & Solution

Explanation:
(a) Since the student cannot see distant objects (like a blackboard at 5 m) clearly, he is suffering from Myopia (short-sightedness).

(b) A Concave (diverging) lens of suitable power is used to correct this defect.

(c) Two main causes of myopia are:
1. Elongation of the eyeball (distance between the eye lens and retina increases).
2. Excessive curvature of the eye lens (focal length of the eye lens becomes too short).

A person suffering from far-sightedness wears a spectacle having a convex lens of focal length 50 cm. What is the distance of the near point of his eye?

Answer & Solution

Answer: $50\text{ cm}$

Explanation: A person with far-sightedness (hypermetropia) cannot see nearby objects clearly. The corrective convex lens takes an object placed at the normal near point ($u = -25\text{ cm}$) and forms a virtual image at the person’s defective near point ($v$).

Given focal length of the convex lens $f = +50\text{ cm}$.

$$\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} – \frac{1}{-25} = \frac{1}{50} \implies \frac{1}{v} + \frac{1}{25} = \frac{1}{50}$$

$$\frac{1}{v} = \frac{1}{50} – \frac{1}{25} = \frac{1 – 2}{50} = -\frac{1}{50}$$

Therefore, $v = -50\text{ cm}$. The defective near point of his eye is $50\text{ cm}$ away.

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Numerical Assignment (Solved)

Class 10 Science · The Human Eye

Eye Defects Practice Set 3

Apply lens formula $\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$ and $P = \frac{1}{f}$

Ch 11 · Human Eye

A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the nature of corrective lens to restore proper vision?

Answer & Solution

Answer: $P = -0.83\text{ D}$, Concave Lens

Explanation: For a myopic eye, the corrective lens must form an image of an object at infinity ($u = -\infty$) at the person’s far point ($v = -1.2\text{ m}$).

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-1.2} – \frac{1}{-\infty} \implies f = -1.2\text{ m}$$

The power of the lens is:

$$P = \frac{1}{f} = \frac{1}{-1.2} \approx -0.833\text{ D}$$

The negative sign indicates that a concave (diverging) lens is required.

The near point of a hypermetropic eye is at 75 cm from the eye. What is the power of the lens required to enable him to read clearly a book held at 25 cm from the eye?

Answer & Solution

Answer: $P = +2.67\text{ D}$ (or 2.66 D)

Explanation: The object to be read is at $u = -25\text{ cm}$. The corrective lens must form a virtual image of this object at the person’s defective near point, so $v = -75\text{ cm}$.

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-75} – \left(-\frac{1}{25}\right) = -\frac{1}{75} + \frac{3}{75} = \frac{2}{75}\text{ cm}^{-1}$$

So, $f = \frac{75}{2}\text{ cm} = 37.5\text{ cm} = +0.375\text{ m}$.

$$P = \frac{1}{f(\text{in m})} = \frac{1}{0.375} = +2.67\text{ D}$$

A myopic person uses specs of power – 0.5 D. What is the distance of far point of his eye?

Answer & Solution

Answer: $2\text{ m}$

Explanation: Given the power of the corrective lens $P = -0.5\text{ D}$. First, calculate the focal length:

$$f = \frac{1}{P} = \frac{1}{-0.5} = -2\text{ m}$$

For a myopic eye, the corrective lens forms a virtual image of an object at infinity at the person’s far point. Therefore, the far point distance is exactly equal to the magnitude of the focal length, which is $2\text{ m}$.

A person wants to read a book placed at 20 cm, whereas near point of his eye is 30 cm. calculate the power of the lens required.

Answer & Solution

Answer: $+1.67\text{ D}$

Explanation: The object (book) is placed at $u = -20\text{ cm}$. The image needs to be formed at the person’s actual near point, so $v = -30\text{ cm}$.

$$\frac{1}{f} = \frac{1}{-30} – \left(-\frac{1}{20}\right) = -\frac{1}{30} + \frac{1}{20} = -\frac{2}{60} + \frac{3}{60} = \frac{1}{60}\text{ cm}^{-1}$$

So, $f = +60\text{ cm} = +0.6\text{ m}$.

$$P = \frac{1}{0.6} = +\frac{10}{6} \approx +1.67\text{ D}$$

The far point distance of a short sighted person is 1.5 meters. find the focal length, power and nature of the remedial lens?

Answer & Solution

Answer: $f = -1.5\text{ m}$, $P = -0.67\text{ D}$, Concave lens

Explanation: To see distant objects clearly, the object is considered at infinity ($u = -\infty$). The image must form at the far point ($v = -1.5\text{ m}$).

$$\frac{1}{f} = \frac{1}{-1.5} – \frac{1}{-\infty} \implies f = -1.5\text{ m}$$

The negative focal length means it is a concave lens.

$$P = \frac{1}{f} = \frac{1}{-1.5} \approx -0.67\text{ D}$$

Eye Defects Practice Set 4

Apply lens formula $\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$ and $P = \frac{1}{f}$

Ch 11 · Human Eye

A person having a myopic eye uses a concave lens of focal length 10 cm. Find the power of the lens.

Answer & Solution

Answer: $-10\text{ D}$

Explanation: Given the focal length of a concave lens is always negative. $f = -10\text{ cm} = -0.1\text{ m}$.

$$P = \frac{1}{f(\text{in m})} = \frac{1}{-0.1} = -10\text{ D}$$

A person with myopic eye cannot see objects beyond 1.2 m distinctly. What should be the nature of corrective lenses to restore proper vision?

Answer & Solution

Answer: Concave Lens, $P = -0.83\text{ D}$

Explanation: Here, distance of far point, $x = 1.2\text{ m}$. For viewing distant objects, the focal length of the corrective lens $f = -x = -1.2\text{ m}$.

$$P = \frac{1}{f} = \frac{1}{-1.2} = -0.833\text{ D}$$

Since power is negative, a concave lens is required.

The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Answer & Solution

Answer: Concave lens, $P = -1.25\text{ D}$

Explanation: To correct myopia, the lens must take an object at infinity ($u = -\infty$) and form an image at the person’s far point ($v = -80\text{ cm} = -0.8\text{ m}$).

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-0.8} – 0 \implies f = -0.8\text{ m}$$

A negative focal length indicates a concave lens.

$$P = \frac{1}{f} = \frac{1}{-0.8} = -1.25\text{ D}$$

The far point of myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to enable him to see very distant objects distinctly?

Answer & Solution

Answer: Concave lens, $P = -1.25\text{ D}$

Explanation: This is solved identically to the previous problem. $v = -80\text{ cm} = -0.8\text{ m}$ and $u = -\infty$.

$$f = -0.8\text{ m}$$

$$P = \frac{1}{f} = \frac{1}{-0.8} = -1.25\text{ D}$$

The far point of a myopic person is 150 cm in front the eye. Calculate the focal length and power of a lens required to enable him to see distant objects clearly.

Answer & Solution

Answer: $f = -1.5\text{ m}$, $P = -0.67\text{ D}$

Explanation: The person is myopic with a far point $v = -150\text{ cm} = -1.5\text{ m}$. To see distant objects clearly, the object distance is $u = -\infty$.

$$\frac{1}{f} = \frac{1}{-1.5} – \frac{1}{-\infty} \implies f = -1.5\text{ m}$$

$$P = \frac{1}{-1.5} \approx -0.667\text{ D}$$