Question

9. The focal length of a concave mirror is 30 cm. find
the position of an object in front of the mirror so that
the image is three times the size of the object.​

Answer 1

Question :-

The focal length of a concave mirror is 30 cm. find the position of an object in front of the mirror so that the image is three times the size of the object.

Answer :-

→ Position of object is at 40 cm ,and image is at 120 cm

To Find :-

→ Position of object .

Explanation :-

Given that :

• Focal length (f) = 30 cm

• size of image = 3 × size of object
• (v = 3 u)
• u = ?

Solution :-

Using the mirror formula -

$\to \sf \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \\ \to \sf \frac{1}{30} = \frac{1}{3u} + \frac{1}{u} \\ \\ \to \sf \frac{1}{30} = \frac{1 + 3}{3u} \\ \\ \sf\to \: \frac{1}{10} = \frac{4}{u} \\ \\ \to \sf \: u \: = 10 \times 4 \\ \\ \to \boxed{ \sf u = 40 \: cm \: }$

hence , position of object (u) = 40 cm

•°• V = 3u

→ v = 120 cm

Answer 2

Answer :

$\bold{Position\:of\:object = 40\ cm }$

Question:

The focal length of a concave mirror is 30 cm. Find the position of object in front of the mirror so that the image is three times the size of object.

Solution:

Given:

• Focal length (f) = 30 cm
• Size of image(v) = 3 × size of object
• Position of object (u) = ?

Atq,

$\longrightarrow\sf v = 3u ...........(i)$

$\rule{300}{1}$

$\sf We \: know \: mirror\: formula :-$

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

$\sf Putting\:value\:from\:(i)\: :$

$\longrightarrow\sf \dfrac{1}{30} =\dfrac{1}{3u} + \dfrac{1}{u} \\\\\longrightarrow\tt \dfrac{1}{30} = \dfrac{1 + 3}{3u} \\\\\longrightarrow\sf \dfrac{1}{30} =\dfrac{4}{3u} \\\\\longrightarrow\sf 3u = 4 \times 30 \\\\\longrightarrow\sf 3u = 120 \\\\\longrightarrow\sf u = \cancel{\dfrac{120}{3}} \\\\\longrightarrow{\boxed{\sf\red{u = 40 \: cm }}}$

$\therefore\sf {Position\: of \: object = 40 \: cm }$

$\therefore\sf {Size \: of \: image = (4\times 30) = 120\: cm }$

Therefore,

Position of object is 30 cm from the mirror.