Question

2. An object of height 2 cm is placed at a distance of
15 cm from a concave mirror of focal length 10 cm.
Draw a scale diagram to locate the image. From the
diagram, find the length of the image formed.
​

Answer 1

Solution:-

Given:-

$\to \rm object \: height \: (h_{o}) = 2cm$

$\to \rm \: focal \: length(f) \: = 10cm$

$\to \rm \: object \: \: distance(u) = 15 \: cm \:$

To find

$\rm \: length \: of \: the \: image \: or \: height \: of \: image \: (h_{i})$

Now using sign conventions

:- focal length is always negative in Concave mirror

$\to \rm \: focal \: length(f) \: = - 10cm$

:- object is opposite to Incident ray

$\to \rm \: object \: \: distance(u) = - 15 \: cm \:$

Formula

$\to \rm \: \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$

Now we get

$\to \rm \: \dfrac{1}{ - 10} = \dfrac{1}{ - 15} + \dfrac{1}{v}$

$\rm \: \to \dfrac{ - 1}{10} + \dfrac{1}{15} = \dfrac{1}{v}$

$\rm \to \: \dfrac{1}{v} = \dfrac{ - 3 + 2}{ 30 }$

$\rm \: \to \: \dfrac{1}{v} = \dfrac{ - 1}{30}$

$\rm \to \: v = - 30 \: cm \:$

We have to find image height

$\rm \: m = \dfrac{ - v}{u} = \dfrac{h_{i}}{h_{o} }$

Now

$\to \rm \: \dfrac{ - ( - 30)}{ - 15} = \dfrac{x}{2}$

$\rm \: \to \: - 2 \times 2 = x$

$\rm \: h _{i} = - 4cm$

So length of image is - 4cm

Answer 2

1

= 30−3+2

1

= 30−1

→v=−30cm

We have to find image height

Now

−15−(−30)

= 2x

2 = x→−2×2=x

=−4cm

So length of image is - 4cm