Question

class 10


A 5cm tall object is placed at a distance of 30cm from a convex mirror of focal length 15cm. Find the postion,size and nature of the image.

Answer 1

hey mate!

here is ur answer,

Explanation:

for a convex mirror,

height of object =5cm

u=30cm

f=15cm

as per sign convention rules

u=-30 cm

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\frac{1}{15} = \frac{1}{v} - \frac{1}{30}$

$\frac{1}{v } = \frac{1}{10} = > v = 10cm$

m=-v/u

=-10/-30=0.3

=hi/ho=hi/5

hi=1.5 cm

the image formed is virtual erect and diminished

hope the answer helps

Answer 2

➤Given:

• Height of the object(h) = 5cm
• Object distance from the convex mirror( u) = -30cm
• Focal length between convex mirror (f) = +15cm

$\rule{400}1$

➤ To Find:

• Position of image
• Size ( magnification )
• Nature of the image form by the mirror.

$\rule{400}1$

➤Solution :

To find Position of the image. We have to use the formula:

$\displaystyle{\sf{\boxed{\red{ \frac{1}{v}+ \frac{1}{u} = \frac{1}{f}}}}}$

:$\implies$ $\displaystyle{\sf{ \frac{1}{v} + \frac{1}{-30} = \frac{1}{15}}}$

:$\implies$ $\displaystyle{\sf{ \frac{1}{v} = \frac{1}{15} - \frac{1}{-30}}}$

:$\implies$ $\displaystyle{\sf{ \frac{1}{v} = \frac{3}{30}}}$

:$\implies$ $\displaystyle{\sf{ \frac{1}{v} = +10}}$

Hence, the position of image is 30cm right side from the pole of convex mirror.

$\rule{400}2$

To find magnification(size) We have to use the formulae :

$\displaystyle{\sf{\boxed{\red{magnificent = \frac{height\:of\:image(-v)}{height\:of\:object(u)}}}}}$

: $\implies$$\displaystyle{\sf{\frac{-(+10)}{(-30)}}}$

: $\implies$$\displaystyle{\sf{\frac{+1}{3}}}$

: $\implies$$\displaystyle{\sf{ \frac{height\:of\:the\:image}{5} = \frac{1}{3}}}$

: $\implies$$\displaystyle{\sf{ \frac{5}{3}cm = 1.67cm}}$

Hence, the virtual image formed by the convex mirror and the size of image is smaller than the size of object.

$\rule{400}4$