 # Human Eye and Colourful World Numerical Problems with Answers

Here you will find numerical problems based on Class 10 Science Chapter 11 Human Eye and Colourful World.

## Numerical Problems for Human Eye and Colourful World

Question 1: A person suffering from the eye-defect myopia (short-sightedness) can see clearly only up to a distance of 2 metres. What is the nature and power of lens required to rectify this defect?

The formula for calculating the power of a lens is: $P = \frac{1}{f}$ Where: $$P$$ = Power of the lens (in diopters, D)
$$f$$ = Focal length of the lens (in meters, m)
In the case of a concave lens used to correct myopia, we have:
Far point of the myopic eye ($$D$$) = 2 meters

The object kept at infinity can be seen clearly if the image of this object is formed at 2 meters. So, the object distance ($$u$$) is infinity ($$\infty$$).

Now, to calculate the focal length of the concave lens required, we can use the lens formula: $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$ Where: $$f$$ = Focal length of the lens (in meters, m)
$$v$$ = Image distance (in meters, m)
$$u$$ = Object distance (in meters, m)

Since the image is formed at 2 meters (the far point of the myopic eye), we have:
$$v = 2$$ meters
And as mentioned earlier, $$u = \infty$$.
So, plugging these values into the lens formula:
$\frac{1}{f} = \frac{1}{-2} – \frac{1}{\infty}$ Since $$\frac{1}{\infty} = 0$$, the equation simplifies to: $\frac{1}{f} = \frac{1}{2} – 0$ $\frac{1}{f} = \frac{1}{-2}$ Now, solving for $$f$$: $f = -2$ Therefore, the focal length ($$f$$) of the concave lens required to rectify myopia is 2 meters.
To find the power ($$P$$) of the lens, we can use the formula $$P = \frac{1}{f}$$: $P = \frac{1}{2}$ So, the power of the concave lens required is $$-0.5$$ diopters (D).

Question 2: The near-point of a person suffering from hypermetropia is at $$50 \, \text{cm}$$ from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is $$25 \, \text{cm}$$)

Convex lens is needed to rectify the defect.
Its calculation for the power of the lens is given by
The defected eye can see the nearby object kept at $$25 \, \text{cm}$$
clearly if the image is formed at its own near point i.e. $$50 \, \text{cm}$$.
Object distance, $$u = -25 \, \text{cm}$$
Image distance, $$v = -50 \, \text{cm}$$
\begin{aligned} & \frac{1}{v} – \frac{1}{u} = \frac{1}{f} \\ & \frac{1}{-50} – \frac{1}{-25} = \frac{1}{f} \\ & f = 50 \, \text{cm} \\ & P = \frac{100}{f} = \frac{100}{50} = 2 \, \text{D} \end{aligned}

Question 3: A person needs a lens of power, – 5.5 dioptres for correcting his distant vision.
For correcting his near vision, he needs a lens of power, +1.5 dioptres. What is the focal length of the lens required for correcting:
(i) distant vision, and (ii) near vision?

(i) For distant vision:
\begin{aligned} P & = -5.5 \, \text{D} \\ P & = \frac{1}{f} \\ f & = \frac{1}{P} = \frac{1}{-5.5} = -0.1818 \, \text{m} \end{aligned}
(ii) For Near Vision:
\begin{aligned} P & = 1.5 \, \text{D} \\ P & = \frac{1}{f} \\ f & = \frac{1}{P} = \frac{1}{1.5} = 0.6666 = 66.66 \, \text{cm} \end{aligned}

Question 4: A person having short-sight cannot see objects clearly beyond a distance of 1.5 m. What would be the nature and power of the corrective lens to restore proper vision?

Far point of the myopic eye ($$D$$) = 1.5 m The object kept at infinity can be seen clearly if the image of this object is formed at 1.6 m.
So, object distance $$u = \infty$$
Image distance, $$v = 1.5 \, \text{m}$$ \begin{aligned} \frac{1}{v} – \frac{1}{u} &= \frac{1}{f} \\ \frac{1}{-1.5} – \frac{1}{\infty} &= \frac{1}{f} \\ f &= -1.5 \, \text{m} \\ P &= \frac{1}{f} = \frac{1}{-1.5} = 0.67 \, \text{D} \end{aligned} So, the nature of the corrective lens needed to restore proper vision is concave, and its power is 0.67 diopters (D).
\begin{aligned} \text{Power} & = \frac{1}{\text{focal length}} \\ \text{Power} & = \frac{1}{-2} \\ \text{Power} & = -0.5 \, \text{D}. \end{aligned} So, the nature of the lens needed to increase the far point to infinity for this short-sighted eye is a concave lens, and its power is -0.5 diopters (D).