A person has been using spectacles of power -1.0 dioptre for distant vision and a separate reading glass of power 2.0 dioptres. What is the least distance of distinct vision for this person :

Solution

Physics sir · 4 steps

The reading glass takes a book at 25 cm and puts its image at this eye's own near point.

1. What each lens tells. The -1.0 D lens fixes the far point, so it plays no part here. The +2.0 D reading glass fixes the near point. It lets the person read a book at the normal 25 cm.
2. Reading glass data. Its focal length is f = 1/2.0 m = +50 cm. The book is the object, so u = -25 cm. Its virtual image must sit at the person's near point.

   1/v = 1/f + 1/u = 1/50 - 1/25 = -1/50

3. Read the image. v = -50 cm. The image is virtual and forms 50 cm in front of the lens, on the same side as the book. The eye sees this image sharply, so that point is its near point.
4. So. The least distance of distinct vision for this person is 50 cm, option (a).

The idea to keep. A corrective lens moves the object to where the eye can see it. For reading glasses, put the object at 25 cm. The virtual image then lands at the eye's true near point.

Answer from the NTA final answer key. Solution by PrepWiser.