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Question: A man wearing glasses of power \(+2D\), can read a book clearly placed at a distance of \(40\;cm\) f...

A man wearing glasses of power +2D+2D, can read a book clearly placed at a distance of 40  cm40\;cm from the eye. The power of the lens required, so that he can read at 25  cm25\;cm from the eye is?
A. +4.5D+4.5D
B. +4.0D+4.0D
C. +3.5D+3.5D
D.+3.0D+3.0D

Explanation

Solution

Power of a lens is the ability or the extent to which the given lens can bend the light ray. According to the nature of the lens, the power either converges or diverges the incident ray. It is measured in m−1m^{-1} or DD.
Formula used:
P=1fP=\dfrac{1}{f} and 1v−1u=1f\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Complete step-by-step solution:
Since power is the ability to bend or focus the incident light to a given point of interest, it is mathematically expressed as
P=1fP=\dfrac{1}{f} where, ff is the focal length of the lens.
For a convex or converging lens, the power is positive and for a concave or diverging lens, the power is negative.
Here, given that power of lens is +2D+2D, which means that the given lens is a concave lens with focal length ff given as
f=1P=1002=0.5mf=\dfrac{1}{P}=\dfrac{100}{2}= 0.5 m
Converting the focal length into cm, we have from the image

f=0.5×100=50cmf=0.5\times 100=50 cm
Since the book is at u=40cmu=40cm. Then from lens law, we have
1v−1u=1f\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}
  ⟹  1v=150−140\implies \dfrac{1}{v}=\dfrac{1}{50}-\dfrac{1}{40}
  ⟹  1v=4−5200\implies \dfrac{1}{v}=\dfrac{4-5}{200}
  ⟹  1v=−1200\implies \dfrac{1}{v}=\dfrac{-1}{200}
  ⟹  v=−200cm\implies v=-200 cm
Now to read the same book from u=25cmu=25cm, the power of the lens required is due to the focal length f′f\prime. Then using the lens law again we have
1v−1u=1f′\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f\prime}
  ⟹  1−200+125=1f′\implies \dfrac{1}{-200}+\dfrac{1}{25}=\dfrac{1}{f\prime}
  ⟹  −1+8200=1f′\implies \dfrac{-1+8}{200}=\dfrac{1}{f\prime}
  ⟹  1f′=7200\implies \dfrac{1}{f\prime}=\dfrac{7}{200}
Here the focal length is in terms of cm, and hence converting it into meter, we get power in terms of dioptre as
∴P=1f′=7200×100=3.5D\therefore P=\dfrac{1}{f\prime}=\dfrac{7}{200}\times 100=3.5D
Thus, the correct answer is C. +3.5D+3.5D

Note: More the power of the lens, more is the refracting ability of it to bend the light, and similarly, less the power of the lens, less is its refracting ability. For convex lens, it is the ability to converge and for concave lens, it is the ability to diverge the incident ray.