Question

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm.
Find the position and nature of the image .​

Answer 1

Answer:

Given: Object distance, u=−10cm Focal length, f=15cm

To find: Image distance, v

From mirror formula :

v

1

+

u

1

=

f

1

v

1

−

10

1

=

15

1

v

1

=

15

1

+

10

1

=

150

25

v=6cm

Hence image will be formed 6 cm beyond the mirror, i.e virtual magnification

m=

u

−v

=

−10

−6

=0.6

Answer 2

Given :

▪ Distance of object = 10cm

▪ Focal length = 15cm

▪ Type of mirror : convex

To Find :

▪ Position and nature of the image.

Concept :

➡ X-coordinates of centre of curvature and focus of concave mirror are negative and those for convex mirror are positive. In case of mirrors since light rays reflect back in X-direction, therefore -ve sign of v indicates real image and +ve sign of v indicates virtual image.

⚽ Mirror Formula :

$\bigstar\:\underline{\boxed{\bf{\pink{\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}}}}}$

⚽ Lateral Magnification :

$\bigstar\:\underline{\boxed{\bf{\purple{m=-\dfrac{v}{u}}}}}$

Calculation :

↗ Position of image :

$\dashrightarrow\sf\:\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}\\ \\ \dashrightarrow\sf\:\dfrac{1}{(-10)}+\dfrac{1}{v}=\dfrac{1}{15}\\ \\ \dashrightarrow\sf\:\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{10}\\ \\ \dashrightarrow\sf\:\dfrac{1}{v}=\dfrac{2+3}{30}=\dfrac{5}{30}\\ \\ \dashrightarrow\underline{\boxed{\bf{\gray{v=6\:cm}}}}\:\orange{\bigstar}$

↗ Nature of image :

$\implies\sf\:m=-\dfrac{v}{u}\\ \\ \implies\sf\:m=-\dfrac{(+6)}{(-10)}\\ \\ \implies\underline{\boxed{\bf{\green{m=0.6}}}}\:\orange{\bigstar}$

• Virtual
• Erect
• Small