Question

The image of an object placed at 10 cm
from the mirror is formed at 10 cm behind the
mirror. If the object is displaced by 4 cm towards
the mirror, by what distance will the image be
displaced with respect to the (i) mirror (ii) object?

Answer 1

Answer:

The distances are taken as positive if measured along the direction of incidence of light and taken as negative if measured along the direction opposite to that of incidence of light.

Given:

Object distance from the mirror, u = -10\ cmu=−10 cm .

Image distance from the mirror, v = +10\ cmv=+10 cm .

Let ff be the focal length of the mirror.

Using mirror formula,

\begin{gathered}\dfrac 1f = \dfrac 1v + \dfrac 1u\\=\dfrac{1}{10}+\dfrac{1}{-10}\\=0\\\Rightarrow f = \infty.\end{gathered}

f

1

=

v

1

+

u

1

=

10

1

+

−10

1

=0

⇒f=∞.

Since, f=\inftyf=∞ , therefore the mirror is plane mirror.

Therefore, when the object is displaced by 4 cm towards the mirror,

The new object distance is u'=-6\ cmu

′

=−6 cm .

Let the image is formed at distance v'v

′

from the mirror.

Using mirror equation,

\begin{gathered}\dfrac 1f = \dfrac {1}{u'}+\dfrac {1}{v'}\\\dfrac {1}{v'}=\dfrac 1f -\dfrac {1}{u'}\\=\dfrac{1}{\infty}-\dfrac{1}{-6} = 6\ cm.\end{gathered}

f

1

=

u

′

1

+

v

′

1

v

′

1

=

f

1

−

u

′

1

=

∞

1

−

−6

1

=6 cm.

Thus, the image is displaced 4 cm with respect to the mirror and 8 cm with respect to the object.