Question

6. An object of size 4cm is placed at a distance of 15 cm from a concave
mirror of focal length 10cm. Find the position, size and nature of image
formed. If the object is moved towards the mirror, what change in size of
the image can be observed?​

Answer 1

Given:-

• Object distance (u) = -15cm
• Focal length (f) = - 10cm
• Height of object (ho) = 4cm

To find:-

• Position, size and nature of image.
• What change in size of the image can be observed when the object is moved towards the mirror.

Solution :-

By using mirror formula,

$\sf \implies \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \\ \\ \sf \implies \: \dfrac{1}{v} = \dfrac{1}{ - 10} - ( \dfrac{1}{ - 15} ) \\ \\ \sf \implies \: \frac{1}{v } = \frac{ - 1}{10} + \frac{1}{15} \\ \\ \sf \implies \frac{1}{v} = \frac{ - 15 + 10}{150} \\ \\ \sf \implies \: \frac{1}{v} = \frac{ - 1}{30} \\ \\ \sf \implies \: \boxed{\boxed{ \bf{v = - 30cm}}}$

∴ Images distance is -30cm.

Now we need to find magnification.

$\sf \implies \: m = \dfrac{ - v}{u} = \dfrac{ h_{i} }{ h_{o} } \\ \\ \sf \implies \frac{ - ( - 30)}{ - 15} = \frac{h_{i}}{4} \\ \\ \sf \implies \: h_{i} = \frac{ - 30 \times 4}{15} \\ \\ \sf \implies h_{i} = \frac{ - 120}{15} \\ \\ \sf \implies \: \boxed{ \boxed{ \bf{h_{i} = - 8cm}}}$

∴ Image hight is -8cm.

As the height of image is negative.

Thus, The image is real, magnified and inverted.

-30cm size of the image can be observed.

Answer 2

According to sign convention ,

Object size (ho) = 4 cm

Object distance (u) = -15 cm

Focal length (f) = -10 cm

We know that , the mirror formula is given by

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

Thus ,

-1/10 = 1/v - 1/15

1/v = -1/10 + 1/15

1/v = (-15 + 10)/150

1/v = -5/150

v = -150/5

v = -30 cm

The negative sign of v shows the image is real and formed on left side of the mirror

Now ,

The magnification of mirror is given by

$\boxed{ \tt{ m =\frac{hi}{ho} = \frac{v}{u} }}$

Thus ,

hi/4 = -(-30)/(-15)

hi/4 = -2

hi = -8 cm

The negative sign of hi shows the image is inverted

Therefore ,

Height of the image (hi) = 8 cm

Position of image (v) = 30 cm

Nature of image = real and inverted