Question

An object is placed 24 cm from a concave mirror. Its
image is inverted and double the size of the object.
Find the focal length of the mirror and the position
where the image is formed.
​

Answer 1

Explanation:

the focal length of the mirror and theposition = 24 + 24 = 48 cm

Answer 2

Given :

In a concave mirror,

Object distance = -24cm.

Height of image = 2 × height of object.

To find :

The focal length and the position of the image.

Solution :

Note : In concave mirror focal length and image distance is negative.

We know that,

» The Magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and is also equal to the ratio of height of the image to the height of the object .i.e.,

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

where,

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

According to question,

• u = -24cm
• h' = 2h

Now, substituting all the given values in the formula,

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

$\dashrightarrow{\sf -\dfrac{ - v}{( - 24)} = \dfrac{2h}{h}}$

$\dashrightarrow{\sf -\dfrac{v}{24} = 2}$

$\dashrightarrow{\sf v = - 2 \times 24}$

$\dashrightarrow{\sf v = - 48 \: cm}$

thus, the position image is -48 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

now, substituting all the given values,

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

$\dashrightarrow{\sf \dfrac{1}{f} = \dfrac{1}{( - 48)} + \dfrac{1}{( - 24)} }$

$\dashrightarrow{\sf \dfrac{1}{f} = - \dfrac{1}{48} - \dfrac{1}{24} }$

$\dashrightarrow{\sf \dfrac{1}{f} = \dfrac{ - 1 - 2}{48}}$

$\dashrightarrow{\sf \dfrac{1}{f} = \dfrac{ - 3}{48}}$

$\dashrightarrow{\sf \dfrac{1}{f} = \dfrac{ - 1 }{16}}$

$\dashrightarrow{\sf f = - 16 \: cm}$

thus, the focal length of the mirror is -16cm.