An object placed 2cm in front of a converging lens of focal length 10cm produces a magnified image of 8cm high determine the position of the image and the size of the object

Answer 1

To determine the position of the image and the size of the object, we can use the magnification formula:

Magnification (M) = - (image height / object height) = (image distance / object distance)

Given data:
Focal length (f) = 10 cm
Object distance (u) = -2 cm
Image height = 8 cm

Since the object distance is negative, it indicates that the object is located on the same side as the incident light. The focal length is positive for a converging lens.

Using the magnification formula, we have:

M = - (image height / object height)

- (8 cm / object height) = (image distance / object distance)

Using the lens formula:

1/f = 1/v - 1/u

1/10 = 1/v + 1/2

1/v = 1/10 - 1/2

1/v = 1/10 - 5/10

1/v = -4/10

v = -2.5 cm

Since the image distance is negative, it indicates that the image is formed on the same side as the object (virtual image).

Now, we can calculate the magnification:

- (8 cm / object height) = (-2.5 cm / -2 cm)

8 / object height = 1.25

object height = 8 / 1.25

object height = 6.4 cm

Therefore, the position of the image is 2.5 cm on the same side of the object, and the size of the object is 6.4 cm.