Question

An object is placed at a distance of 10cm from convex mirror and an image is formed 6 cm behind the convex mirror. What is its magnification?​

Answer 1

Answer:

6 cm

Explanation:

Given that

u= -10 cm

f= 15 cm

Find out

We need to find v

Formula

We know the formula

1/f = 1/v + 1/u

Therefore, 1/v= 1/f – 1/u

Substituting the known values in the above equation we get,

1/v= 1/15 + 1/10

1/v = (2+3)/30 (lcm)

Hencev=5/30 = 6 cm

v= 6cm

Nature of the image: The image is formed behind the mirror and it is virtual and erect

Answer 2

$\maltese{\underline{\underline{\bold{\sf{Given\:that: —}}}}}$

An object is placed at a distance of 10cm from convex mirror.

An image is formed 6cm behind the convex mirror.

$\maltese{\underline{\underline{\bold{\sf{To\:Find: —}}}}}$

What is its magnification?

$\maltese{\underline{\underline{\bold{\sf{Using\:Formula: —}}}}}$

Formula for finding the focal length —

${\qquad}{\bigstar}{\boxed{\bold{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f}}}}$

For finding the magnification of the image —

${\qquad}{\bigstar}{\boxed{\bold{m = \frac{ - v}{u}}}}$

$\maltese{\underline{\underline{\bold{\sf{Full\:Solution: —}}}}}$

$\bigstar$According to the Question —

${\longrightarrow} {\tt{ \frac{1}{6} - \frac{1}{10} = \frac{1}{f}}} \\ \\ {\longrightarrow} {\tt{ \frac{5 - 3}{30} = \frac{1}{f}}} \\ \\ {\longrightarrow} {\tt{\cancel\frac{2}{30} = \frac{1}{f}}} \\ \\ {\longrightarrow} {\tt{f = 15 \: cm}}$

Hence focal length will be formed 15 CM beyond the mirror,i.e virtual magnification

${\longrightarrow} {\tt{m = \frac{ - 6}{ - 10}}} \\ \\ {\longrightarrow} {\tt{m = 0.6}}$

Hence,it will be erect and diminished.