Question

Draw the image formation by a concave mirror of focal length 15cm for the following

positions of the object. Also, indicate the nature and relative size of the image:

(i)The object is placed at 30cm from the mirror.

(ii) The object is placed at 10cm from the mirror.​

Answer 1

Answer:

hope this will help you

Answer 2

Answer:

(i)Real inverted and of the same size as the object.

(ii)virtual, erect, and of the same size as the object.

Explanation:

We will use the mirror formula to solve this question. Which is given as,

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

Where,

f=focal length of the mirror=-15cm             (given)

v=image distance from the mirror

u=object distance from the mirror

(i)The object distance(u₁)=-30cm

By inserting the required values in equation (1) we get;

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

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

$\mathrm{\frac{1}{v}= \frac{-2+1}{30}}$

$\mathrm{\frac{1}{v}= \frac{-1}{30}}$

$\mathrm{v=-30\ cm}$                 (2)

The magnification of the mirror is calculated as,

$\mathrm{m=\frac{-v}{u} }$         (3)

By inserting the values in equation (3) we get;

$\mathrm{m=\frac{-(-30)}{-30} }$

$\mathrm{m=-1}$                (4)

From equations (2) and (4) we can say that the image formed is real inverted and of the same size as the object.

(ii)The object distance(u₂)=-10cm

By inserting the required values in equation (1) we get;

$\mathrm{\frac{1}{-15} =\frac{1}{v}+ \frac{1}{-10}}$

$\mathrm{\frac{1}{v}= \frac{-1}{15}+\frac{1}{10} }$

$\mathrm{\frac{1}{v}= \frac{-2+3}{30}}$

$\mathrm{\frac{1}{v}= \frac{1}{30}}$

$\mathrm{v=30\ cm}$                   (5)

By inserting the values in equation (3) we get;

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

$\mathrm{m=1 }$                              (6)

From equations (5) and (6) we can say that the image formed is virtual, erect, and of the same size as the object.

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