### Chapter 1: Light - Reflection and Refraction

Q
##### Light - Reflection and Reflection

Question:

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Here, the given mirror concave.

Object Distance ($u$) = -27 cm

Focal length ($f$) = -18 cm

Height of object ($h_o$) = 7 cm

Image Distance ($v$) = ?

Height of image ($h_i$) = ?

Object distance ($u$) is taken negative as its distance from the pole of the mirror is measured opposite to the direction of the incident light.

By using the mirror formula

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

$\frac{1}{-18} = \frac{1}{v}+\frac{1}{(-27)}$

$\frac{1}{-18} = \frac{1}{v} - \frac{1}{27}$

$\frac{1}{-18} +\frac{1}{27} = \frac{1}{v}$

$\frac{1}{v} = \frac{-3 + 2}{54}$

$\frac{1}{v} = \frac{-1}{54}$

$v = -54 cm$

The screen should be placed at a distance of 54 cm in front of the mirror to obtain a sharp image.

Now, Magnification
$m = \frac{Height\quad of \quad image (h_i)}{ Height\quad of \quad object (h_o)} = \frac{size\quad of \quad image (v)}{size\quad of \quad object (u)}$
$m=\frac{h_i}{h_o}= -\frac{v}{u}$

$m= -\frac{v}{u}$

$m= -\frac{-54}{-27}$

$m= -2$

Here, the magnification is negative and more than 1. The negative sign indicates that the image is inverted and value more than 2 indicates that the size of the image is larger than the object.

Now,

$m = \frac{h_i}{h_o}$

$-2 = \frac{h_i}{7}$

$h_i = -2 \times 7 = -14 cm$

The negative sign of image height $h_i\$ indicates that the image is inverted.

Therefore, the image is real, inverted and enlarged in size.

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