Question

A man wearing glasses of $3D$must hold a newspaper at least $25cm$ away to see the print clearly. How far would the newspaper have to be if he took off the glasses and still wanted clear vision?
(A). $0.4m$ away from his eyes
(B). $0.5m$ away from his eyes
(C). $0.6m$ away from his eyes
(D). $1m$away from his eyes

Answer 1

Hint: Since the power is positive, the lens used is a convex lens. The power is the reciprocal of focal length, so we can calculate the focal length. Using the lens formula and substituting corresponding values we can find the image distance which is the minimum distance at which objects must be kept so that the man can see clearly without glasses.
Formulas used:
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
$f=\dfrac{1}{P}$

Complete answer:
It is clear that the man is far-sighted, this means he suffers from Hypermetropia and cannot see nearby objects clearly. The lens of a certain power helps by creating a virtual image of the object at a distance where it is clearly visible to the man.
Therefore, the least distance at which the man must hold the newspaper is $25cm$ therefore, the least distance where he can see objects clearly can be calculated using the lens formula-
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ - (1)
Here, $v$ is the image distance
$u$ is the object distance
$f$is the focal length
For the condition given above, $u=-25cm$, the focal length is
$\begin{align}
  & f=\dfrac{1}{P} \\
 & \Rightarrow f=\dfrac{100}{3}cm \\
\end{align}$
We substitute given values in eq (1) to get,
$\begin{align}
  & \dfrac{1}{v}+\dfrac{1}{25}=\dfrac{3}{100} \\
 & \Rightarrow \dfrac{1}{v}=\dfrac{3}{100}-\dfrac{1}{25} \\
 & \Rightarrow \dfrac{1}{v}=\dfrac{-1}{100} \\
 & \therefore v=-100cm=-1m \\
\end{align}$
The minimum distance at which the man can see objects is $1m$.
Therefore, in order for the man to see objects clearly without glasses, the minimum distance at which objects must be kept is $1m$.

Hence, the correct option is (D).

Note:
By convention, all distances from left to right are taken as negative and distances from right to left are taken as positive. Therefore, object distance is always negative. Here, image distance is negative because the image formed is virtual. To correct Hypermetropia, convex lenses are used.