Question

The near point and the far point of a child are at 10cm and 100cm. If the retina is 2cm behind the eye lens. What is the range of the power of the eye lens?
A. 50D to 40D
B. 60D to 51D
C. 60D to 54D
D. 50D to 40D

Answer 1

Use the lens formula and find the focal lengths of the eye lens for the near point and far point. Then use the relation between the power of the lens and its focal length. Be careful while substituting the value of focal length while finding the power.
Formula used:
$P=\dfrac{1}{f}$
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Complete answer:
We see the objects that are in front of us because of our eyes. The eye lens present in our eye helps in focusing the rays of light on the retina in our eye. The eye lens has the ability to adjust its focal length.
To see an object, the light rays from the object must focus on the retina. This means that the image of the object must be formed on the retina.
Let us now understand what is meant by near point and far point of vision.
Near the point is the minimum distance of the object from us, so that we can see the object clearly.
Far point is the maximum distance of the object, until where the object is seen clearly.
Therefore, for an object to be seen clearly, it must be between the near and far points.
Power is defined as the reciprocal of the focal length, where the focal is in metres.
i.e. $P=\dfrac{1}{f}$.
Therefore, let us find the focal lengths of the lens for the near point and far point. For this, we will use the lens formula $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ ….. (i).
Here, v and u are the positions of the image and the object.
Let us find the focal length (${{f}_{1}}$) for the near point.
In this case, v = 2cm and u = -10cm .
$\Rightarrow \dfrac{1}{{{f}_{1}}}=\dfrac{1}{2}-\dfrac{1}{-10}=\dfrac{6}{10}=\dfrac{3}{5}$
$\Rightarrow {{f}_{1}}=\dfrac{5}{3}cm=\dfrac{5}{3}\times {{10}^{-2}}m$.
${{P}_{1}}=\dfrac{1}{{{f}_{1}}}=\dfrac{3}{5}\times 100=60D$.
Let us find the focal length (${{f}_{2}}$) for the far point.
In this case, v = 2cm and u = -100cm .
$\Rightarrow \dfrac{1}{{{f}_{1}}}=\dfrac{1}{2}-\dfrac{1}{-100}=\dfrac{51}{100}$
$\Rightarrow {{f}_{1}}=\dfrac{100}{51}cm=\dfrac{1}{51}m$.
${{P}_{1}}=\dfrac{1}{{{f}_{1}}}=51D$.
Therefore, the range of the power of the eye lens is 60D to 51D.

Hence, the correct option is B.

Note:
Note that the unit of power of a lens is dioptre (D). 1D = 1${{m}^{-1}}$. Therefore, the value of the focal length that we substitute to find the value of the power must be in metres. Students may sometimes do this incorrectly.