Question

An object is placed 0.18 m away from a concave mirror whose focal length is 0.1m. Find the position and size of the image of the object size is 0.04m.

Answer 1

Answer:

The image will form 0.22m in front of the mirror

and the size of the image is 0.048m , it will be enlarged and real image .

Explanation:

Let us consider the following

• the object distance be u

• the image distance be v

• the focal length be f

• the magnification be m

Formulae to be used

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

and

$m = \frac{ h_{2}}{ h_{1} } = - \frac{v}{u} \: \\ where \: \pink{h_{1} } \: and \: \pink{h_{2} } \: are \: the \: sizes \: of \\ and \: image \: respectively$

Given data is :

u = -0.18m

f = -0.1m

h₁ =0.04m

Using mirror formula we have

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \implies \frac{1}{v} + \frac{1}{ - 0.18} = \frac{1}{ -0 .1} \\ \implies \frac{1}{v} = - \frac{10}{1} + \frac{100}{18} \\ \implies \frac{1}{v} = \frac{10 \times 18 + 100}{18} \\ \implies \frac{1}{v} = \frac{ - 180 + 100}{18} \\ \implies \frac{1}{v} = \frac{ - 80}{18} \\ \implies v = - \frac{18}{80} \\ \implies v = - 0.225m$

Therefore , the image distance is -0.22m i.e. the image is formed 0.22m in front of the image . (negative sign signify that the image is real and forms in front of the image)

Now we know that

$m = \frac{h_{2} }{h_{1}} = - \frac{v}{u}$

So using these values

$\frac{h_{2}}{h_{1}} = - \frac{v}{u} \\ \implies \frac{h_{2}}{0.04} = - \frac{0.22}{0.18} \\ \implies h_{2}= \frac{22}{18} \times 0.04 \\ \implies h_{2} = 1.22 \times 0.04 \\ \implies h_{2} = 0.048m$

The image size is 0.048m .

Thus image is enlarged

Answer 2

Answer:

v=0.225

and

h2= 0.048

Explanation:

1/v+1/u=1/f

1/v+1/-0.18=1/-0.1

1/v=-10/1+100/18

1/v=10*8+100/18

1/v=-80/18

v=-18/80

v=0.225m