### Chapter 1: Light - Reflection and Refraction

Q
##### Light - Reflection and Reflection

Question:

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Here, the given mirror convex.

Object Distance ($u$) = -20 cm

Radius of curvature ($R$) = +30 cm

Height of object (${h}{o}$) = 5 cm

Image Distance ($v$) = ?

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

Using the relation between $R$ and $f$:

$R = 2f$

We get,

$f = \frac{R}{2}$

$f = \frac{30}{2}$

$f = 15 cm$

Now, using the mirror formula

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

$\frac{1}{15} = \frac{1}{v}+\frac{1}{(-20)}$

$\frac{1}{15} = \frac{1}{v} - \frac{1}{20}$
$\frac{1}{15} +\frac{1}{20} = \frac{1}{v}$

$\frac{1}{v} = \frac{4-3}{60}$

$\frac{1}{v} = \frac{7}{60}$

$v=\frac{60}{7}= 8.57 cm$

The positive sign of $v$ indicates that the image is formed on the other side of the mirror at a distance of 8.57 cm. That is the image formed is virtual.

Now, magnification

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

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

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

$m= -\frac{8.57}{-20} = 0.428$

Here, the magnification is positive and less than 1. The positive sign indicates that the image is erect and value less than 1 indicates that the size of the image is smaller than the object.$m = \frac{h_i}{h_o}$
$0.428 = \frac{h_i}{5}$

$h_i = 0.428 \times 5 = 2.14 cm$

The positive sign of image height $h_i$ indicates that the image is erect.

Therefore, the image is formed at a distance of 0.428 cm on the other side of the mirror. The image formed is 2.14 cm tall and it is virtual, erect and smaller in size.

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