Question

Q. i) An object of height 8 cm is placed at a distance of 25 cm in front of a lens having radius of curvature 30 cm The image of an object is obtained on the screen. Determine the nature of lens used. Also find the nature, size and position of image formed.
(ii). A person cannot see the objects beyond 600 cm. Find the power of lens he should use to correct this defect.

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Answer 1

Question 1:

Given:

object of height 8 cm is placed at a distance of 25cm in front of the lens with radius of curvature 30cm.

To find:

Nature , size and position of the image

Calculation:

Applying Lens Formula ;

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

$= > \: \dfrac{1}{15} = \dfrac{1}{v} - \dfrac{1}{( - 25)}$

$= > \: \dfrac{1}{15} = \dfrac{1}{v} + \dfrac{1}{25}$

$= > \: \dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{25}$

$= > \: \dfrac{1}{v} = \dfrac{5 - 3}{75}$

$= > \: \dfrac{1}{v} = \dfrac{2}{75}$

$= > v = 37.5 \: cm$

Let image size be d ;

$= > \: \dfrac{d}{8} = \dfrac{37.5}{ - 25}$

$= > d = (- 1.5) \times 8$

$= > d = - 12 \: cm$

So, image is real inverted and magnified.

Question 2:

Let focus of lens be f ;

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

$= > \: \dfrac{1}{f} = \dfrac{1}{ - 600} - \dfrac{1}{ ( - \infty)}$

$= > \: \dfrac{1}{f} = \dfrac{1}{ - 600} - 0$

$= > \: \dfrac{1}{f} = \dfrac{1}{ - 600}$

$= > f - 600 \: cm$

$= > power = \dfrac{100}{ - 600} = - 0.167 \: dioptre$

So, the lens used is a concave lens of power 0.167 D