Question

an object 5 cm high is placed at a distance of 10cm from a convex mirror of radius of curvature 40cm find the nature ,position and size of the image.​

Answer 1

Solution:-

Here distance, u = -10 cm (to left of mirror)

Image distancev= ? (to be calculated)

And,

focal length,f= Radius of curvature /2

= 40/2 cm

= +20cm (convex mirror)

putting these values in mirror formula :-

1/v +1/u =1/f

we get,

1/v +1/(-10) =1/+20

or

1/v-1/10= 1/20

1/v = 1/20 + 1/10

1/v = 1+2/20

1/v = 3/20

so, image distance, v = +0.15cm

Thus, the position of image is 0.15cm behind the convex Mirror. since the image is formed behind the convex mirror, its nature will be virtual and erect.

Now,

Magnification, m = -v/u

Here, image distance, v = +0.15cm

And Object distance, u = -10cm

So, m = (+0.15) /(-10)

m= 0.15/10

Magnification, m = 0.015

we also have another formula for magnification, which is :-

m =h2/h1

Here, magnification, m = +0.015

height of image, h2 = ?

and height of object, h1= +5cm

So, + 0.015 = h2/(+5)

h2 = 5*0.015

Height of image, h2 = 0.075cm

thus, the size of image is 0.075cm.

Answer 2

Answer:

Explanation:

The properties of the image is virtual,diminished,errect

1/u+1/v=1/f

-1/10+1/v=1/15

1/v=3+2/30

1/v=1/6

v=6

According to the linear magnification

Height of the image/Height of the object

-6/10

.6

Again

.6=Height of the image/5

Height of the image is 3.0cm that is 3cm