Question

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. ​

Answer 1

Given:

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 is obtained on the screen.

To find:

• Type of lens

• Nature , Size and position of image

Calculation:

Radius of curvature = 30 cm ;

Focal length = 30/2 = 15 cm.

Since focal length is positive , the lens is converging (i.e. Convex lens).

Applying Lens Formula ;

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

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

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

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

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

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

$\boxed{ = > v = 37.5 \: cm}$

Magnification be m ;

$m = \dfrac{v}{u} = \dfrac{37.5}{ - 25} = - 1.5$

$= > \: | m| > 1$

So, nature of image :

• Image is real , inverted and magnified.

Hope It Helps.