Question

An object 5cm in length is held 30 cm away from a converging lens of

focal length 10cm. Draw the ray diagram and find the position, size and nature

of the image formed.​

Answer 1

Given:

       length of object = l = 5cm

       focal length = f = 10cm

       position of object =  u = -25 cm

To find :

    1) position of image = v =?

    2) height of image = ?

    3) nature of image = ?

Step-by-step explanation:

1)

       By lens formula:

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

        $\displaystyle \frac{1}{v} = \frac{1}{u} + \frac{1}{f}$

      $\displaystyle \frac{1}{-25} + \frac{1}{10}$

       $\displaystyle { \frac{1}{v} = \frac{3}{50} }$

       $v = \frac{50}{3}$

    v = 16.66cm

The distance of the image is 16.66 on the opposite side of the lens.

2)

     Magnification = $\displaystyle \frac{16.66}{-25}$

     m = -0.66

     Also, we have

     $\displaystyle m= \frac{height\ of\ image}{height \ of \ object}$

      $\displaystyle -0.66 =\frac{h}{5}$

     h = 5 $\times$ (-0.66)

   h = -3.3 cm

The negative sign shows that the image of the object is formed inverted.

    h = 3.3 cm

3)

    Nature of image = real & inverted

Refer attached figure here: