(a) A person suffering from myopia (near-sightedness) was advised to wear corrective lens of power - 2·5 D. A spherical lens of same focal length was taken in the laboratory. At what distance should a student place an object from this lens so that it forms an image at a distance of 10 cm from the lens?(b) Draw a ray diagram to show the position and nature of the image formed in the above case.

(a) Given:

Power of lens, $P=-2.5D$

Image distance, $v=-10cm$

To find: Object distance from the lens, $u$.

Solution: 

From the magnification formula, we know that-

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

Putting the value of power in the formula we get-

$-2.5=\frac {1}{f}$

$f=-\frac {1}{2.5}$

$f=-\frac {10}{25}$

$f=-0.4m,\ or\ -40cm$

Thus, the focal length, $f$ of the lens is 40cm.

Now, 

From the lens formula, we know that-

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

Putting the required values, we get-

$\frac {1}{(-10)}-\frac {1}{u}=\frac {1}{(-40)}$

$-\frac {1}{10}-\frac {1}{u}=-\frac {1}{40}$

$-\frac {1}{10}+\frac {1}{40}=$

$\frac {1}{u}=\frac {-4+1}{40}$

$\frac {1}{u}=-\frac {3}{40}$

$u=-\frac {40}{3}$

$u=-13.3cm$

Thus, the object distance, $u$ is 13.3cm in front of the lens.

(b) As the power of the lens is $-ve$, the lens used is concave/diverging. 

The position of the image is on the same side as that of the object (on the left), and the nature of the image is virtual and erect.

A ray diagram given below to show the position and nature of the image formed in the above case.

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Updated on: 10-Oct-2022

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