Question

when an object placed at 30 cm in fornt of concave mirror image of the same size is formed what is the focal length of the image​

Answer 1

Given :

In concave mirror,

Object distance = 30 cm.

Height of object = height of the image.

To find :

The focal length of the image.

Solution :

1st we have to find image distance, 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,

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

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,

$\dashrightarrow\sf \dfrac{h}{h} = - \dfrac{v}{ - 30}$

$\dashrightarrow\sf 1 = \dfrac{v}{30}$

$\dashrightarrow\sf v = 30 \: cm$

Thus, the position of the image is 30 cm

Now 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,

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

$\dashrightarrow\sf \dfrac{1}{ - 30} + \dfrac{1}{ - 30} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{1}{ - 30} - \dfrac{1}{ 30} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{ - 1 - 1}{ 30} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{ - 2}{ 30} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{1}{f} = \dfrac{ - 1}{15}$

$\dashrightarrow\sf f = - 15 \: cm$

Thus, the focal length of the image is 15 cm.

Answer 2

Explanation:

Given:-

In a concave mirror

• object distance =u=-30cm
• Height of the object(h)=Heightof the image (h')

To find:-

Focal length of the mirror =f

Solution:-

We know that

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

$\\ \tt{:}\Rrightarrow 1=-\dfrac{v}{-30}$

$\\ \tt{:}\Rrightarrow 1=\dfrac{v}{30}$

$\\ \tt{:}\Rrightarrow v=30cm$

According to mirror formula

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

$\\ \tt{:}\Rrightarrow \dfrac{1}{-30}+\dfrac{1}{-30}=\dfrac{1}{f}$

$\\ \tt{:}\Rrightarrow \dfrac{-1+(-1}{30}=\dfrac{1}{f}$

$\\ \tt{:}\Rrightarrow \dfrac{-1-1}{30}=\dfrac{1}{f}$

$\\ \tt{:}\Rrightarrow \dfrac{-2}{30}=\dfrac{1}{f}$

$\\ \tt{:}\Rrightarrow \dfrac{-1}{15}=\dfrac{1}{f}$

$\\ \tt{:}\Rrightarrow f=-15cm$