Question

An object is placed at a distance of 10 cm in front of a concave mirror with a focal length of 15 cm. What is the position and the nature of the image formed?

Answer 1

Given :

• An object is placed at a distance of 15 cm

• Focal length = 10 cm.

To find :

• We have to find the image distance ?

Solution :

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.

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.