Question

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)

Answer 1

A person with a defective eye-vision is unable to see the objects nearer than 1.5 m. He wants to read books at a distance of 30 cm. Find the nature, focal length and power of the lens he needs in his spectacles.

This person suffers from the defect of hypermetropia.

For him u = -30cm, v = -1.5 m = -150cm

Therefore, focal length of corrective lens to be used by him is

1/f = 1/v- 1/u = 1/-150 - 1/-30 = 4/150 = 37.5cm

The positive sign shows that the lens needed is a convex lens of focal length 37.5 cm.

Hence, power of lens needed

P =1/f = 100/37.5 = 2.67D

Answer 2

The focal length and power of the lens is 0.375 m and 2.67 D.

Explanation:

Given that,

Object distance = 30 cm = 0.3 m

Image distance = -1.5 m

The image distance will be taken to be negative because the image is formed at a distance less than nearest point

We need to calculate the focal length

Using formula of focal length

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-1.5}+\dfrac{1}{0.3}$

$\dfrac{1}{f}=\dfrac{8}{3}$

$f=0.375\ m$

Therefore, The focal length is positive so the lens will be convex lens.

We need to calculate the power

Using formula of power

$P=\dfrac{1}{f}$

$P=\dfrac{1}{0.375}$

$P=2.67\ D$

Hence, The focal length and power of the lens is 0.375 m and 2.67 D.

Learn more :

Topic : Focal length

https://brainly.in/question/12933655