Question

an object placed on a metre scale at 8m mark was focused on a white screen placed at 92cms, using a converging lens placed on the scale at 50cms mark.
-find 'f' of converging lens
-find position of image formed if object is shifted towards the lens at the position of 29cms
-state nature of image formed if the object is further shifted towards the lens.

Answer 1

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
⇒1/v + 1/21 = 1/21
⇒ 1/v = 1/21 -1/21 = 0
⇒ v = ∞

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.



,

Answer 2

Answer:

Explanation:(i) Given: object is placed at 8cm.

Real image is formed at 92 cm mark.

Therefore, object distance u=-42 cm

Applying the lens formula:

(ii) Focal length of lens f =+21 cm

Object distance u = -21 cm

Applying the lens formula:

v=infinity

(iii)If the object is further shifted towards the lens, the image formed will be virtual, erect and magnified.