Question

An object is placed on a meter scale at 8cm marked was focus on a white screen placed at 92cm marked using a converging lens placed on the scale at 50cm marked. Find the focal length, image distance if the object shifted to the lens at position of 29cm. State the nature of image formed if the object is further shifted to the lens

Answer 1

1.

Object distance, u = -(50 – 8) = -42 cm

Image distance, v = (92 – 50) = 42 cm

Now,

1/f = 1/v – 1/u

=> 1/f = 1/42 + 1/42

=> f = 21 cm

2.

When the object is shifted to position 29, the new object distance is, u = -(50 – 29) = -21 cm

That means the object is placed at the focus. So, the image will be formed at infinity.

1/f = 1/v – 1/u

=> 1/21 = 1/v + 1/21

=> v = infinity

3.

If the object is further shifted towards the lens, the object distance becomes less than focal length and it lies between focus and pole of the lens. So, the image formed will be virtual, erect and magnified. The image lies on the same side of the object.

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.