An object is placed 30 cm in front of a converg... - SS2 Physics Geometrical Optics Question

Object distance (u) = -30 cm

Focal length of converging lens (f1) = 15 cm

Focal length of diverging lens (f2) = -20 cm

To find the final image distance (v), we can use the lens formula:

1/f1 = (1/v) - (1/u1),

1/f2 = (1/v) - (1/u2).

Since the image formed by the converging lens serves as the object for the diverging lens, the object distance for the diverging lens (u2) is the negative of the image distance formed by the converging lens.

u2 = -v1.

Substituting these values into the lens formula, we have:

1/15 = (1/v) - (1/-30),

1/-20 = (1/v) - (1/-v1).

Simplifying the equations, we get:

1/v = 1/15 + 1/30,

1/v = 1/20 - 1/v1.

Now, we solve the first equation for 1/v:

1/v = 2/30 + 1/30 = 3/30,

1/v = 3/30,

v = 30/3 = 10 cm.

Substituting the value of v into the second equation, we have:

1/10 = 1/20 - 1/v1.

To isolate 1/v1, we rearrange the equation:

1/v1 = 1/20 - 1/10 = 1/20 - 2/20 = -1/20.

Taking the reciprocal of both sides, we get:

v1 = -20 cm.

Since the image distance v1 is negative, the final image formed by the diverging lens is a virtual image.

Therefore, the final image distance is -20 cm, and it is a virtual image.