Question

Q9.An object 5cm in height is placed at a distance of 20cm in front of a convex
lens of focal length 10cm.Find the height position and nature of the image.
Illustrate your answer with the help of suitable diagram.​

Answer 1

Answer:

Explanation:

Given;

        Convex lens.

        Height of object or $H_{o}$ = 5cm;

       Object distance or u = -20cm;

        Focal length or f = +10cm;

Formula to be applied :

Mirror's Formula :  $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$;

Magnification = +(v/u) =  $H_{i}$/ $H_{o}$;

Image distance equals;

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

                          $\frac{1}{10} = \frac{1}{v} - \frac{1}{(-20)}$;

                          $\frac{1}{10} = \frac{1}{v} - \frac{1}{(-20)}$;

                          $\frac{1}{v} = \frac{1}{10} + \frac{1}{(-20)}$;

                          $\frac{1}{v} = \frac{-2 + 1}{-20}$;

                          $\frac{1}{v} = \frac{-1}{-20}$;

                          $\frac{1}{v} = \frac{1}{20}$;

Thus,                 v = 20 cm;

The image formed will be 20 cm away, on the other side, of the lens. It will form on the center of curvature which is on the other side of the lens.

Height of image or  $H_{i}$  equals;

              v/u =  $H_{i}$/ $H_{o}$;

            20/(-10) =  $H_{i}$/ 5;

             $H_{i}$ = 20*5/(-10);

              $H_{i}$ = -10 cm;

Nature;

•          Real and inverted.
•          Magnified

That's all.