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?
Answer:
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 \(25 \, \text{cm}\))
Answer:
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?
Answer
(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?
Answer:
A person having short-sight cannot see objects clearly beyond a distance of 1.5m.
The nature and power of the corrective lens to restore proper vision is concave. The calculation of the power is shown as follows:
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). Question 5: An eye has a far point of 2m. What type of lens in spectacles would be needed to increase the far point to infinity? Also calculate the power of lens required. Is this eye long-sighted or short-sighted?
Answer: This is a case of short-sightedness. A concave lens of focal length 2m should be used to curb this defect. The power of this lens should be calculated as follows:
\[ \begin{aligned} \text{Power} & = \frac{1}{\text{focal length}} \\ \text{Power} & = \frac{1}{-2} \\ \text{Power} & = -0.5 \, \text{D}. \end{aligned} \] 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).