Question

Make a ray diagram of an object placed at a distance of 60cm from a concave lens of focal length 30cm.

Answer 1

Height of object, h₀ = 4cm

object distance from lens , u = 30cm

Focal length of lens , f = 20cm

Now, use formula,

1/v - 1/u = 1/f

Here, u = -30cm, f = 20cm

∴ 1/v -1/-30 = 1/20

⇒ 1/v = 1/-30 + 1/20 = (30 - 20)/20×30 = 1/60

⇒1/v = 1/60 ⇒ v = 60cm

now, we should use formula of magnification

m = v/u = height of image/height of object

Here , v = 60cm, u = -30cm, height of object = 4cm ,

so, 60cm/-30cm = height of image/4cm

-2 = height of image/4

height of image = -8cm , here negative sign shows image is formed below of optical axis { horizontal line }

Now, size of image/size of object = 8cm/4cm = 2 { excluding sign }

Hence, height of image or size of image = 8cm

image distance form lens = 60cm , right side

and ratio of image size or object size is 2 {excluding sign }

ray diagram of image , its position , principal focus are shown in figure.

Where h shown height of image e.g., 8cm , v is shown distance of image from lens e.g., 60cm

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Answer 2

Given,

U=-30cm, m=1/2

Using the formula m= -v/u

½= -v/-30

V=+15cm.

Using the mirror formula,

1/f=1/v+1/u

1/f= 1/15+1/-30

F=+30 cm.

Again m’ =-v’/u’

1/3 =-v’/u’

V’ =-u’/3

Using again, 1/f= 1/v’+1/u’

1/30=-3/u’+1/u’

1/30= -2/u’

U’=-60cm.