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Question: A far-sighted person cannot focus distinctly on objects closer than 120 cm. The lens that will permi...

A far-sighted person cannot focus distinctly on objects closer than 120 cm. The lens that will permit him to read from a distance of 40 cm will have a focal length:
A) +30cm+ 30\,cm
B) 30cm- 30\,cm
C) +60cm+ 60\,cm
D) 60cm- 60\,cm

Explanation

Solution

Since we have been given the image and object distance, we can use the lens formula to find the focal length of the lens. We will require a lens that can diverge the light coming from the object to a distance of 120 cm.

Formula used: In this solution, we will use the following formula:
Lens formula: 1v1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} where vv is the image distance, uu is the object distance, and ff is the focal length of the lens

Complete step by step answer
We’ve been told that a far sighted person cannot focus on objects closer than 120 cm. So, let’s assume in our scenario that the image is formed at a distance of 120 cm from the person. The focal length of the lens must be such that the object when placed at a distance of 40 cm from the person should form an image at 120 cm.
So, we can say that u=40cmu = - 40\,cm and v=120cmv = - 120\,cm. Then using the lens formula, we can write
1120140=1f\Rightarrow \dfrac{1}{{ - 120}} - \dfrac{1}{{ - 40}} = \dfrac{1}{f}
Taking the LCM on the left side, we get
1120+3120=1f\Rightarrow \dfrac{1}{{ - 120}} + \dfrac{3}{{ - 120}} = \dfrac{1}{f}
Which gives us,
1f=2120\Rightarrow \dfrac{1}{f} = - \dfrac{2}{{120}}
f=60cm\therefore f = - 60\,cm

Hence the focal length of the lens must be 60cm- 60\,cm which corresponds to option (D).

Note
Since we want the image to be formed behind the object, we would want to use a diverging lens which has a negative focal length so we can rule out options (A) and (C) directly. Here we have neglected the focal length of our eye and assumed that the only lens at play is the external lens.