Question

An object is placed 20cm in front of a convex mirror and the image is formed 10 cm from the

mirror. Calculate new image distance if object is moved 5cm towards the mirror​

Answer 1

• Distance of object from mirror = 20cm

• Distance of image from mirror = 10cm

Let focal length of mirror be f.

• According to the question :-

$\rm \longrightarrow\dfrac{1}{10} + \dfrac{1}{( - 20)} = \dfrac{1}{f} \\ \\ \\ \rm \longrightarrow\dfrac{1}{10} - \dfrac{1}{20} = \dfrac{1}{f} \\ \\ \\ \rm \longrightarrow \dfrac{2 - 1}{20} = \dfrac{1}{f} \\ \\ \\ \rm \longrightarrow \dfrac{ 1}{20} = \dfrac{1}{f} \\ \\ \\ \\ \\ \rm \longrightarrow \: 20 \: cm = f$

Therefore, focal length of mirror = 20 cm

Now, if object is moved 5cm towards the mirror then distance of object from mirror = 20cm - 5 cm

= 15 cm

Let new image distance be v.

According to the question :-

$\rm \longrightarrow \dfrac{1}{v} + \dfrac{1}{( - 15)} = \dfrac{1}{20} \\ \\ \\ \rm \longrightarrow \dfrac{1}{v} = \dfrac{1}{20} + \dfrac{1}{ 15}\\ \\ \\ \rm \longrightarrow \dfrac{1}{v} =\dfrac{3 + 4}{60}\\ \\ \\ \rm \longrightarrow \dfrac{1}{v} =\dfrac{7}{60}\\ \\ \\ \rm \longrightarrow {v} =\dfrac{60}{7}cm$

Therefore , new image distance = 60/7 = 8.57 cm

Answer 2

Explanation:

[tex]

Distance of object from mirror = 20cm

Distance of image from mirror = 10cm

Let focal length of mirror be f.

According to the question :-

\begin{gathered} \rm \longrightarrow\dfrac{1}{10} + \dfrac{1}{( - 20)} = \dfrac{1}{f} \\ \\ \\ \rm \longrightarrow\dfrac{1}{10} - \dfrac{1}{20} = \dfrac{1}{f} \\ \\ \\ \rm \longrightarrow \dfrac{2 - 1}{20} = \dfrac{1}{f} \\ \\ \\ \rm \longrightarrow \dfrac{ 1}{20} = \dfrac{1}{f} \\ \\ \\ \\ \\ \rm \longrightarrow \: 20 \: cm = f\end{gathered}

⟶

10

1

+

(−20)

1

=

f

1

⟶

10

1

−

20

1

=

f

1

⟶

20

2−1

=

f

1

⟶

20

1

=

f

1

⟶20cm=f

Therefore, focal length of mirror = 20 cm

Now, if object is moved 5cm towards the mirror then distance of object from mirror = 20cm - 5 cm

= 15 cm

Let new image distance be v.

According to the question :-

\begin{gathered} \rm \longrightarrow \dfrac{1}{v} + \dfrac{1}{( - 15)} = \dfrac{1}{20} \\ \\ \\ \rm \longrightarrow \dfrac{1}{v} = \dfrac{1}{20} + \dfrac{1}{ 15}\\ \\ \\ \rm \longrightarrow \dfrac{1}{v} =\dfrac{3 + 4}{60}\\ \\ \\ \rm \longrightarrow \dfrac{1}{v} =\dfrac{7}{60}\\ \\ \\ \rm \longrightarrow {v} =\dfrac{60}{7}cm\end{gathered}

⟶

v

1

+

(−15)

1

=

20

1

⟶

v

1

=

20

1

+

15

1

⟶

v

1

=

60

3+4

⟶

v

1

=

60

7

⟶v=

7

60

cm

Therefore , new image distance = 60/7 = 8.57 cm

[tex]