Question

A object 5 cm in length is placed at distance 20 cm in frpnt of convex mirror

Answer 1

*Answer:*

Position of Image : Behind the mirror.

Nature of Image : Virtual and Erect.

Size of the Image: Diminished.

*Explanation:*

Given:

Height of object = 5 cm

Object Distance (u) = - 20 cm (By using Sign Convention)

Radius of Curvature (R) = 30 cm

Focal Length (f) = 2R = 30 / 2 = 15 cm (By using Sign Convention)

Using Mirror Formula:

{\boxed{\boxed{\sf{1/v + 1/u = 1/f}}}}

1/v+1/u=1/f

1/v - 1/20 = 1/15

1/v = 1/15 + 1/20

1/v = 4 + 3 / 60 (Taking LCM of 15 and 20)

1/v = 7 / 60

Image distance is 8.57 cm behind the convex mirror. As the v is positive the image formed is virtual and erect.

{\boxed{\boxed{\sf{Magnification = - v / u }}}}

Magnification=−v/u

= - 8.57 / - 20 = 0.4 cm

This indicates that the image is diminished because the magnification of the image is less than 1.

{\boxed{\boxed{\sf{Magnification = Height Of Object / Height Of Image}}}}

Magnification=HeightOfObject/HeightOfImage

= hi / ho = 0.4

We know ho here,

hi / 5 = 0.4

hi = 0.4 * 5

hi = 2 cm

As the image formed is smaller than the height of the object the image formed is diminished.

Therefore the final answer,

Position of Image : Behind the mirror.

Nature of Image : Virtual and Erect.

Size of the Image: Diminished.