Question

an object is placed at a distance of 8 cm from the convex mirror of focal length 12 cm find the position of the image formed​

Answer 1

Answer:

Explanation:

Given,

Object Distance, u = - 8 cm

Focal length, f = 12 cm

To Find,

Image distance, v = ?

Formula to be used,

Mirror formula,

1/f = 1/v + 1/u

Solution,

Putting all the values, we get

1/f = 1/v + 1/u

⇒ 1/12 = 1/v + 1/- 8

⇒ 1/12 = 1/v + 1/8

⇒ 1/v = 1/8 + 1/12

⇒ 1/v = 3 + 2/24

⇒ 1/v = 5/24

⇒ v = 24/5

⇒ v = 4.8 cm

Hence, the image distance is 4.8 cm.

Here, positive sign refers that the image will be virtual and erect.

Answer 2

Answer:

Given :-

• An object is placed at a distance of 8 cm from the convex mirror of focal length 12 cm.

To Find :-

• What is the position of the image formed.

Formula Used :-

$\clubsuit$ Mirror Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{1}{f} =\: \dfrac{1}{v} + \dfrac{1}{u}}}}$

where,

• f = Focal Length
• u = Object Distance
• v = Image Distance

Solution :-

Given :

$\bigstar\: \: \bf{Object\: Distance =\: -\: 8\: cm}$

$\bigstar\: \: \bf{Focal\: Length =\: 12\: cm}$

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{1}{12} =\: \dfrac{1}{v} + \bigg(- \dfrac{1}{8}\bigg)$

$\longrightarrow \sf \dfrac{1}{12} =\: \dfrac{1}{v} - \dfrac{1}{8}$

$\longrightarrow \sf \dfrac{1}{12} + \dfrac{1}{8} =\: \dfrac{1}{v}$

$\longrightarrow \sf \dfrac{2 + 3}{24} =\: \dfrac{1}{v}$

$\longrightarrow \sf \dfrac{5}{24} =\: \dfrac{1}{v}$

By doing cross multiplication we get,

$\longrightarrow \sf 5v =\: 24(1)$

$\longrightarrow \sf 5v =\: 24$

$\longrightarrow \sf v =\: \dfrac{24}{5}$

$\longrightarrow \sf\bold{\red{v =\: 4.8\: cm}}$

${\small{\bold{\green{\underline{\leadsto\: The\: image\: distance\: is\: 4.8\: cm\: .}}}}}$

${\small{\bold{\purple{\underline{\therefore\: The\: position\: of\: the\: image\: formed\: is\: virtual\: and\: erect\: .}}}}}$