Question

A 2 cm high object is placed at a distance of 32 cm from a concave mirror. The image is real, inverted and 4 cm in size. Find the focal length of the mirror and the position where the image is formed​

Answer 1

Given:-

• Height of the object $\sf{(h_o)}$ = 2 cm
• Object distance (u) = -32 cm
• Image formed is real.
• Height of the image $\sf{(h_i)}$ = -4 cm
• Mirror = Concave

To Find:-

• Focal length of the mirror.
• Position of the image formed.

Solution:-

We know,

The formula of magnification of a mirror is:-

• $\sf{m = \dfrac{h_i}{h_o}}$

Hence, putting the values we have in the magnification formula:-

= $\sf{m = \dfrac{-4}{2}}$

$\sf{\implies m = -2}$

∴ The magnification of the mirror is -2 cm.

Also we know,

Magnification of a mirror is:-

• $\sf{m = \dfrac{-v}{u}}$

Hence,

$\sf{-2 = \dfrac{-v}{-32}}$

$\sf{\implies -v = -32\times -2}$

$\sf{\implies -v = 64}$

$\sf{\implies v = -64}$

∴ The distance of the image from the mirror is -64 cm.

Now,

Applying mirror formula:-

$\sf{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}$

= $\sf{\dfrac{1}{f} = \dfrac{1}{-64} + \dfrac{1}{-32}}$

= $\sf{\dfrac{1}{f} = \dfrac{-1 - 2}{64}}$

= $\sf{\dfrac{1}{f} = \dfrac{-3}{64}}$

= $\sf{f = \dfrac{-64}{3}}$

= $\sf{f = -21.3}$

$\boxed{\bf{\underline{\therefore\:The\: focal\: length\: of \:the \:mirror\: is \:21.3 \:cm}}}$

Now,

Finding the radius of curvature of the mirror.

We know,

Radius of curvature = 2 × Focal Length

= Radius of curvature = 2 = 21.3 × 2 = 42.6 cm

∴ The radius of curvature of the mirror is 42.6 cm

Since the object is at a distance of 32 cm (which comes in between 21.3 cm and 42.6 cm), the object is placed between radius of curvature (C) and focus of the mirror (F)

______________________________________

Answer 2

Given :-

• Height of the object ${\sf{(h_{o}}}$= 2 cm
• Object distance ( u ) = -32 cm
• Image formed is real
• Height of the image ${\sf{(h_{i}}}$
• Mirror = Concave Mirror.

To find :-

• Focal Length of the Mirror.
• Position of the Image formed.

★ As we know that,

$\large\dag$ Formula Used

• $\dashrightarrow$ m = $\large{\sf{\frac{h_{i}}{h_{o}}}}$

$\large\dag$ Hence,

• Putting the value, magnification formula :

• $\longmapsto$m = $\large{\sf{\frac{-4}{2}}}$

• $\longmapsto$m = -2

• The Magnification of the Mirror is = $\large\underline{\rm{-2~cm}}$

★ As we know that,

❍ Magnification of a Mirror is :

• $\large\boxed{\sf{m = \frac{-v}{u}}}$

$\large\dag$ Hence,

:$\implies$-2 = $\large{\sf{\frac{-v}{-32}}}$

$~~~~~$:$\implies$-v = -32 × -2

$~~~~~~~~~~$:$\implies$-v = 64

$~~~~~~~~~~~~~~~$:$\implies$${\underline{\boxed{\pink{\frak{v~=~-64}}}}}$★

$\large\dag$ Henceforth,

• This Distance is the Image from the Mirror is = $\large\underline{\rm{-64~cm}}$

Now,

$\large\dag$ Applying Mirror Formula :

:$\implies$$\large{\sf{\frac{1}{f} = \frac{1}{v} \: + \: \frac{1}{u}}}$

$~~~~~$:$\implies$$\large{\sf{\frac{1}{f} = \frac{1}{-64} \: + \: \frac{1}{-32}}}$

$~~~~~~~~~~$:$\implies$$\large{\sf{\frac{-1}{f}}}$$\large{\sf{\frac{-1 \: -2}{64}}}$

$~~~~~~~~~~~~~~~$:$\implies$$\large{\sf{\frac{1}{f}}}$= $\large{\sf{\frac{-3}{64}}}$

$~~~~~~~~~~~~~~~~~~~~$:$\implies$f = $\large{\sf{\frac{-64}{3}}}$

$~~~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$f = ${\cancel{\dfrac{-64}{3}}}$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$${\underline{\boxed{\pink{\frak{f~=~-21.3}}}}}$★

$\large\dag$ Hence Verified,

• The Focal Length of the Mirror is = $\large\underline{\rm{21.3~cm}}$

Now,

• Finding the Radius of the Curvature of the Mirror.

★ As we know that,

• Radius of the Curvature = 2 × Focal Length.

$\hookrightarrow$ Radius of the Curvature = 2 = 21.3 × 2 = 42.6cm.

.°. The Radius of the Curvature of the Mirror is = $\large\underline{\rm{42.6~cm}}$

Since, the object is distance of 32 cm (which comes in between 21.3 cm and 42.6 cm). The Object is placed between radius of curvature $\large{\bf{(C)}}$ and Focus of the Mirror $\large{\bf{(E)}}$.