Question

09.(2) An object ofheight 5 cm placed in front of concave mimor with focal length 10cm it is placed a 20cm
distance from mirror find size location and nature of image​

Answer 1

Explanation:

As per the provided information in the given question, we have :

• Height of the object $\bf (h_o)$ = 5 cm
• Focal length of the mirror (f) = –10 cm

[ Focal length of concave mirror is always taken negative.]

• Object distance (u) = – 20 cm

[ Object is always placed to left of the mirror and , the distance towards left of the mirror is taken negative as per the sign convention. ]

We are asked to calculate the size, location(position) and the nature of the image.

★ Calculating position or location of the image :

In order to calculate the position or location of the image, we need to find the image distance (v).

By using mirror formula,

$\\ \longrightarrow \quad\pmb{\boxed{ \sf { \dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f} }}}$

• u denotes object distance
• v denotes image distance
• f denotes focal length

$\\ \longrightarrow \quad \sf { \dfrac{1}{(-20)} + \dfrac{1}{v} = \dfrac{1}{(-10)} }$

$\\ \longrightarrow \quad \sf { - \dfrac{1}{20} + \dfrac{1}{v} = -\dfrac{1}{10} }$

$\\ \longrightarrow \quad \sf { \dfrac{1}{v} = -\dfrac{1}{10} +\dfrac{1}{20} }$

$\\ \longrightarrow \quad \sf { \dfrac{1}{v} = \dfrac{1}{20} -\dfrac{1}{10} }$

$\\ \longrightarrow \quad \sf { \dfrac{1}{v} = \dfrac{1-2}{20} }$

$\\ \longrightarrow \quad \sf { \dfrac{1}{v} = \dfrac{-1}{20} }$

Reciprocating both sides,

$\\ \longrightarrow \quad \sf { v= \dfrac{20}{-1} }$

$\\ \longrightarrow \quad \bf \underline { v= -20 \; cm }$

Position of the image : 20 cm to the left side of the mirror.

$\underline{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}$

★ Calculating size of the image :

By using magnification formula,

$\\ \longrightarrow \quad\pmb{\boxed{ \sf { m = \dfrac{h_i}{h_o} }}}$

Also,

$\\ \longrightarrow \quad\pmb{\boxed{ \sf { m = -\dfrac{v}{u} }}}$

• m denotes magnification
• h_o denotes object height
• h_i denotes height of the image
• v denotes image distance
• u denotes object distance

Comparing both equations, we get :

$\\ \longrightarrow \quad \sf { \dfrac{h_i}{h_o}= -\dfrac{v}{u} }$

$\\ \longrightarrow \quad \sf { \dfrac{h_i}{5}= -\dfrac{(-20)}{(-20)} }$

$\\ \longrightarrow \quad \sf { \dfrac{h_i}{5}= -(1) }$

$\\ \longrightarrow \quad \sf { h_i= -1 \times 5}$

$\\ \longrightarrow \quad \bf \underline { h_i = -5 \; cm}$

Size of the image : Image is 5 cm long and it is formed below the principal axis.

$\underline{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}$

Let us find the magnification first.

$\\ \longrightarrow \quad \sf { \dfrac{h_i}{h_o}= m }$

$\\ \longrightarrow \quad \sf { -\dfrac{(-20)}{(-20)} = m}$

$\\ \longrightarrow \quad \sf { -1 = m}$

Nature of the image : Since the magnification has sign of subtraction (as the image is formed below the principal axis), so the image is real and inverted.