Question

A concave lens of focal length 15 cm forms an image 10cm from the lens. Calculate
(i) the distance of the object from the concave lens
(ii) the magnification for the image formed
(iii) the nature of the image formed

Answer 1

according to lens formula

Answer 2

Answer:

(i) The distance of the object from the lens is -30cm.

(ii)The magnification for the image formed is 0.33.

(iii)The nature of the image is virtual, erect, and diminished.

Explanation:

We will use the lens formula to solve this question i.e.

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$                  (1)

Where,

f=focal length of the lens

v=distance of the image from the lens

u=distance of the object from the lens

From the question we have,

v=-10cm              (∴ Negative sign indicates that the image is virtual)

f=-15cm

By substituting the value of f and v in equation (1) we get;

$\frac{1}{-15}=\frac{1}{-10}-\frac{1}{u}$

$\frac{1}{u}=\frac{1}{-10}+\frac{1}{15}=\frac{-3+2}{30}=\frac{-1}{30}$

$u=-30cm$

Magnification(m) of the lens is given as,

$m=\frac{v}{u}=\frac{-10}{-30}=\frac{1}{3}=0.33$          (2)

From the value of the magnification, we can say that the image is erect and diminished.

Hence,

(i) The distance of the object from the lens is -30cm.

(ii)The magnification for the image formed is 0.33.

(iii)The nature of the image formed is virtual, erect, and diminished.