A person with a defective eye vision is unable to see the objects nearer than 1.5 m. He wants to read a book at a distance of 30 cm. Find the focal length and the power of the lens he needs in his spectacles . (Kindly explain that whether we have to take image distance negative or positive)

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Solution

$T h e i m a g e d i s \tan c e w i l l b e t a k e n t o b e - i v e b e c a u s e i n o r d e r f o r t h e p e r s o n t o r e a d t h e i m a g e m u s t b e f o r m e d b e y o n d 1.5 m o r a t t h e v e r y l e a s t a t 1.5 m a w a y i n f r o n t o f h i s e y e s . \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \frac{1}{f} = \frac{1}{- 1.5} - \frac{1}{- 0.3} \frac{1}{f} = \frac{- 1 + 5}{1.5} \frac{1}{f} = \frac{4}{1.5} f = 0.375 m$

Power = 1/f
P = 1/0.375

P = 2.67 Dioptres
The lens will be convex lens.

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A person with a defective eye vision is unable to see the objects nearer than 1.5 m. He wants to read a book at a distance of 30 cm. Find the focal length and the power of the lens he needs in his spectacles . (Kindly explain that whether we have to take image distance negative or positive)

The image distance will be taken to be -ive because in order for the person to read the image must be formed beyond 1.5m orat the very least at 1.5m away infront of his eyes.1f= 1v-1u1f= 1-1.5-1-0.31f= -1+51.51f=41.5f=0.375m Power = 1/f P = 1/0.375 P = 2.67 Dioptres ​The lens will be convex lens.