Question

an object is placed at a distance of 15 cm from a concave mirror of focal length 10 cm find the position and the nature of the image​

Answer 1

Answer:

Explanation:

Given,

Object distance, u = - 15 cm

Focal length, f = 10 cm

To Find,

Image distance, v = ?

Formula to be used,

Mirror Formula,

⇒ 1/v + 1/u = 1/f

Solution,

Putting all the values, we get

1/v + 1/u = 1/f

⇒ 1/v + 1/(- 15) = 1/10

⇒ 1/v - 1/15 = 1/10

⇒ 1/v = 1/10 + 1/15

⇒ 1/v = (3 + 2)/30

⇒ 1/v = 5/30

⇒ 1/v = 1/6

⇒ v = 6 cm

Hence, the image distance is 6 cm.

Nature of the image,

The image is formed behind the mirror and it is virtual and erect.

Answer 2

Explanation:

$\frak Given = \begin{cases} &\sf{An\ object\ is\ placed\ at\ a\ distance\ of\ 15cm.} \\ &\sf{Focal\ length\ =\ 10cm.} \end{cases}$

To find:- We have to find the image distance ?

__________________

$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {Mirror\ formula\ =\ \dfrac{1}{v}\ +\ \dfrac{1}{u}\ =\ \dfrac{1}{f}.}$

Here:-

• v, is for image distance.
• u, is for object distance.
• f, is for focal length.

__________________

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {\dfrac{1}{v}\ +\ \dfrac{1}{(-15)}\ =\ \dfrac{1}{10}} \\ \\ \sf : \implies {\dfrac{1}{v}\ -\ \dfrac{1}{15}\ =\ \dfrac{1}{10}} \\ \\ \sf : \implies {\dfrac{1}{v}\ =\ \dfrac{1}{10}\ +\ \dfrac{1}{15}} \\ \\ \sf : \implies {\dfrac{1}{v}\ =\ \dfrac{(3+2)}{30}} \\ \\ \sf : \implies {\dfrac{1}{v}\ =\ \dfrac{5}{30}} \\ \\ \sf : \implies {\dfrac{1}{v}\ =\ \dfrac{1}{6}} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf v\ =\ 6cm.}}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ required\ image\ distance\ is\ 6cm.}}$

__________________

$\frak{\underline{\underline{\dag Nature\ of\ the\ image:-}}}$

● The required image is formed behind the mirror, hence it is virtual and erect.