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

Step 1: Image formation by the convex lens

When an object is at infinity, a convex lens forms its image at its focal point.

Given focal length of the convex lens $f_1 = 30$ cm.

Size of the image formed by the convex lens $h'_1 = 2$ cm.

The distance of this image from the convex lens is $v_1 = f_1 = 30$ cm.

Step 2: Object for the concave lens

A concave lens of focal length $f_2 = -20$ cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens.

The image formed by the convex lens acts as the object for the concave lens.

The object distance for the concave lens ($u_2$) is the distance between the image formed by the convex lens and the concave lens.

$u_2 = v_1 - 26 = 30 - 26 = 4$ cm.

Since the object is to the left of the concave lens (in the direction of light travel from convex lens image), $u_2$ is positive according to sign convention used in lens formula.

Step 3: Image distance for the concave lens

Using the lens formula for the concave lens:

$\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}$

$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2}$

$\frac{1}{v_2} = \frac{1}{-20} + \frac{1}{4} = \frac{-1 + 5}{20} = \frac{4}{20} = \frac{1}{5}$

$v_2 = 5$ cm.

The image is formed at a distance of 5 cm from the concave lens, and it is positive, meaning it is on the right side of the concave lens (same side as the refracted light).

Step 4: Magnification by the concave lens

Magnification produced by the concave lens is $m_2 = \frac{v_2}{u_2} = \frac{5}{4}$.

Step 5: Size of the final image

The size of the image formed by the convex lens is the object size for the concave lens, so $h_2 = h'_1 = 2$ cm.

The size of the final image formed by the concave lens is $h'_2 = m_2 \times h_2 = \frac{5}{4} \times 2 = \frac{10}{4} = 2.5$ cm.

Final Answer: The final answer is ${2.5 \, \text{cm}}$

Answer 2

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: \[ \text{size of image} = 2.5 \, \text{cm} \]