Question

an object 5cm high is placed at a distance of 10cm from a convex mirror of radius of curvature 45cm ,find the nature position and size of the image physics chapter reflection and refr
action​

Answer 1

Answer:

Nature of Image : Virtual and erect ,

Position of Image : At a dist of 9cm behind the mirror

Size of image = 4.5 cm height (diminished)

Explanation:

height of object h₁ = 5 cm

object distance u = - 10 cm

Radius of curvature R = + 45 cm   ⇒ focal length f = 2R = + 90 cm

From mirror formula

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

$\frac{1}{v}$ - $\frac{1}{10}$ = $\frac{1}{90}$

$\frac{1}{v}$ = $\frac{1}{90}$ + $\frac{1}{10}$

$\frac{1}{v}$ = $\frac{10}{90}$

∴ v = + 9 cm

∴ Image is formed behind the mirror at a dist of + 9 cm

magnification m = -v/u = - 9/(-10) = + 0.9

Since m is positive image is virtual and erect.

Since m < 1, image is diminished.

Also m = h₂/h₁

        Size of image h₂ = m h₁ = 0.9 x 5 = 4.5 cm  

Nature of Image : Virtual and erect ,

Position of Image : At a dist of 9cm behind the mirror

Size of image = 4.5 cm height (diminished)