Question

An object 0.1 m tall is placed 0.4 m from a convex mirror with a focal length of 0.3 m. What is the height
of the image?​

Answer 1

Given :

In convex mirror,

Object height : 0.1 m

Object distance : 0.4 m

Focal length : 0.3 m

To find :

The height of the image.

Solution :

First we have to find the image distance, using mirror formula that is,

» A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

$\boxed{\bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}$

where,

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

By substituting all the given values in the formula,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 0.4} = \dfrac{1}{0.3}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{ 0.4} = \dfrac{1}{0.3}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{0.3} + \dfrac{1}{ 0.4}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{4 + 3}{1.2}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{7}{1.2}$

$\dashrightarrow\sf v= \dfrac{1.2}{7}$

$\dashrightarrow\sf v= 0.1 \: m$

Now,

we know that,

» The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and it is equal to the ratio of image height and object height. that is,

$\boxed{\bf \dfrac{h'}{h} = - \dfrac{v}{u}}$

where,

• h' denotes height of the image
• h denotes height of the object
• v denotes image distance
• u denotes object distance

By substituting all the given values in the formula,

$\dashrightarrow\sf \dfrac{h'}{h} = - \dfrac{v}{u}$

$\dashrightarrow\sf \dfrac{h'}{0.1} = - \dfrac{0.1}{ - 0.4}$

$\dashrightarrow\sf h' = \dfrac{0.01}{0.4}$

$\dashrightarrow\sf h' = 0.025 \: m$

Thus, the height of the image is 0.025 m.

Answer 2

Answer:-

• Object height=0.1m=h
• Object distance=-0.4m=u
• Focal length=f=0.3m
• Image distance=v=?
• Image height=h'=?

We know

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

$\\ \sf\hookrightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\\ \sf\hookrightarrow \dfrac{1}{v}=\dfrac{1}{0.3}-\dfrac{1}{-0.4}$

$\\ \sf\hookrightarrow \dfrac{1}{v}=\dfrac{1}{0.3}+\dfrac{1}{0.4}$

$\\ \sf\hookrightarrow \dfrac{1}{v}=\dfrac{4+3}{1.2}$

$\\ \sf\hookrightarrow \dfrac{1}{v}=\dfrac{7}{1.2}$

$\\ \sf\hookrightarrow v=\dfrac{1.2}{7}$

$\\ \sf\hookrightarrow v=0.1m$

Now.

$\boxed{\sf Magnification=\dfrac{h'}{h}=-\dfrac{v}{v}}$

$\\ \sf\hookrightarrow \dfrac{h'}{0.1}=-\dfrac{0.1}{-0.4}$

$\\ \sf\hookrightarrow \dfrac{h'}{0.1}=\dfrac{0.1}{0.4}$

$\\ \sf\hookrightarrow h'=\dfrac{0.01}{0.4}$

$\\ \sf\hookrightarrow h'=0.025$