Question

an object is kept at a distance of 30 cm from a diverging lens of focal length 15 CM at what distance the image is formed from the lens find the magnification of the image?​

Answer 1

Answer:

Substitute f=−30cm and v=−15cm to find the value of u. Substitute, v=−15cm, u=−30cm and h1=5cm to find the value of h2. Therefore, the object should be placed 30 cm from the optical centre and the size of the image formed is 2.5 cm.

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Answer 2

Given :

Object distance (u) = - 30 cm

Focal length (f) = - 15 cm

To Find :

(i) Image distance

(ii) Magnification of the image

Solution :

(i) Using Lens formula,

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

⇒  -$\frac{1}{15}$   = - $\frac{1}{30}$ + $\frac{1}{v}$

⇒  $\frac{1}{v}$     =  $\frac{1}{30}$ - $\frac{1}{15}$

⇒  $\frac{1}{v}$    =  $\frac{1 - 2}{30}$

∴  v     = - 30 cm

Hence the image is formed at the same side as the object at a distance of 30 cm from the lens.

(ii) Magnification (m) = $\frac{v}{u}$

                                  = $\frac{-30}{-30}$

                                  = 1

Therefore, the magnification of the image is 1.