Question

Two convex lenses A and B, each of focal length 10.0 cm, are mounted on an optical bench at 50.0 cm and 70.0 cm respectively. An object is mounted at 20.0 cm. Find the nature and position of the final image formed by the combination.

Answer 1

Step 1: First Lens (Lens A at 50.0 cm).

Focal length (f_A) = 10.0 cm

Object distance (u_A) = 20.0 − 50.0

= −30.0 cm    ...[Since the object is before the lens, it is negative]

Using the lens formula:

`1/v_A = 1/f_A + 1/u_A`

= `1/10 - 1/30`

= `(3 - 1)/30`

= `2/30`

`1/v_A = 1/15`

v_A = 15 cm

So, the image formed by Lens A is at 50 + 15 = 65 cm on the optical bench.

Step 2: Second Lens (Lens B at 70.0 cm).

The image formed by Lens A (at 65 cm) acts as the object for Lens B.

Focal length (f_B) = 10.0 cm

Object distance (u_B) = 65 − 70.0

= −5.0 cm    ...[Negative since it’s before the lens]

Using the lens formula:

`1/v_B = 1/f_B + 1/u_B`

= `1/10 - 1/5`

= `(1 - 2)/10`

`1/v_B = (-1)/10`

v_B = −10 cm

The negative v_B indicates that the final image is virtual and located 10 cm in front of Lens B.

Lens B is at 70 cm, so the final image is at:

70 − 10 = 60 cm

∴ The image is upright and magnified.