Question

An object placed on a metre scale at 30 cm mark was focused on a white screen at 70 cm mark, using a converging lens placed on the scale at 50 cm mark.Find the focal length of converging lens
Find the position of the image formed if the object is shifted towards the lens at position of 40 cm.State the nature of the image formed if the object is further shifted towards the lens.​

Answer 1

Answer:

An object placed on a metre scale at 30 cm mark was focused on a white screen at 70 cm mark, using a converging lens placed on the scale at 50 cm mark.Find the focal length of converging lens

Find the position of the image formed if the object is shifted towards the lens at position of 40 cm.State the nature of the image formed if the object is further shifted towards the lens.

Answer 2

Answer:

1.

position of object 8 cm

position of lens = 50 cm

position of image = 92 cm

object distance is the distance between

lens and object = 50-8 = 42 cm

image distance is the distance between

lens and image = 92-50 = 42 cm

So u = -42cm, v= +42 cm, f=?

We know that when the object is at the centre of curvature, the object and image distance are the same. radius of curvature = object distance = 42 cm. So focal length = half of radius of curvature = 42/2 = 21cm. Or you can use the formula to get the focal length as

1/v-1/u = 1/f

1/42 - 1/(-42) = 1/f 1/42 + 1/42 = 1/f 1/f = 2/42 = 1/21

= f= 21cm

2.

If the object is shifted to 29cm new object distance =50-29 = 21 cm or u=-21cm

Now the object is at focus, so the image is formed at infinity. Or you can calculate by formula as

1/v 1/(-21) = 1/21

3 1/v + 1/21 = 1/21

1/v = 1/21 -1/21 = 0

* V= 00

3.

Now if the object is further shifted towards the lens, the object will be between focus and optical centre. So the image will be virtual, erect and magnified.