Question

An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror

radius of curvature 30 cm find the position, nature and size of image.​

Answer 1

Position =
`(60)/(7)\ cm` behind the mirror

Nature = Virtual and Erect

Size =
`(15)/(7)\ cm` : Diminished

Step-by-step explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

`(1)/(u)+(1)/(v)=(1)/(f)\\ (1)/(-20)+(1)/(v)=(1)/(15)\\ (1)/(v)=(7)/(60)\\v=(60)/(7)\ cm`

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification =
`=(h_(image))/(h_(object))=`
`-(v)/(u)=(60)/(7*20)=(3)/(7)`

Height of the object = 5 cm

Height of the image =
`5*(3)/(7)=(15)/(7)\ cm`

Since the height of the image is positive and less than the size of object,it is erect and diminished.