Question

The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the lens, the new size of the image is :

• A. 1.25 cm
• B. 2.5 cm
• C. 1.05 cm
• D. 2 cm

Answer 1

The Correct Option is B

The initial image size is given as: \[ I_1 = 2 \, \text{cm} \].

This image \( I_1 \) serves as a virtual object for the concave lens, which then forms a new image at \( I_2 \).

Using the lens formula: \[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]

For the concave lens:

\[\frac{1}{v} - \frac{1}{4} = -\frac{1}{20}\]

Solving for \( v \):

\[\frac{1}{v} = -\frac{1}{20} + \frac{1}{4}\]

\[\frac{1}{v} = \frac{-1 + 5}{20} = \frac{4}{20} = \frac{1}{5}\]

Thus, the image distance is: \[ v = 5 \, \text{cm} \] (distance of \( I_2 \) from the concave lens).

Calculating the magnification:

\[\text{Magnification} = \frac{v}{u} = \frac{\text{size of image}}{\text{size of object}} = \frac{5}{4} = 1.25\]

Therefore:

\[\frac{\text{size of image}}{2} = 1.25\]

The final image size produced by the concave lens is: \[ \text{size of image} = 2.5 \, \text{cm} \]