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

To find the new size of the image when a concave lens is placed in the system, we need to consider how this additional lens affects the image formed by the convex lens alone.

Step 1: Image formation by the convex lens 

The object is placed at infinity, and a convex lens of focal length \(30 \, \text{cm}\) is used. The image size formed by this convex lens is given as \(2 \, \text{cm}\).

Thus, the convex lens forms an image at its focal length on the opposite side since the object is at infinity.

Step 2: Effect of the concave lens

The concave lens is placed \(26 \, \text{cm}\) from this image and has a focal length of \(-20 \, \text{cm}\) (negative for concave lenses).

We will calculate the position of the final image using the lens formula:

\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

Here, \(u = 26 \, \text{cm}\) (the distance from the concave lens to the initial image), and \(f = -20 \, \text{cm}\).

Substituting these into the formula, we have:

\(\frac{1}{v} - \frac{1}{26} = \frac{1}{-20}\)

Solving for \(v\) gives:

\(\frac{1}{v} = \frac{1}{26} - \frac{1}{20}\)

\(\frac{1}{v} = \frac{20 - 26}{520}\)

\(\frac{1}{v} = -\frac{6}{520}\)

\(v = -\frac{520}{6} \approx -86.67 \, \text{cm}\)

The negative sign indicates that the image is formed on the same side as the object for the concave lens, which is virtual.

Step 3: Magnification and new image size

The magnification \((m)\) induced by the concave lens is given by:

\(m = \frac{v}{u} = \frac{-86.67}{26} \approx -3.33\)

The overall magnification of the system is the product of the magnifications by the convex lens and the concave lens:

\(m_{\text{total}} = 1 \times (-3.33)\)

(Since the initial image size was 2 cm, the magnification of the convex lens itself is effectively 1 here due to the object's position at infinity.)

The new image size would be:

2.5 \, \text{cm}.

Answer 2

$A 'B'$ is the real image due to convex lens and it is at Focus of convex lens. $A'B'$ acts as virtual object for concave lens and object distance is $+4 \,cm$ $\frac{1}{f} =\frac{1}{v}-\frac{1}{u} v=\frac{20}{4} $ $\frac{1}{v} =\frac{1}{f}+\frac{1}{u} v=5 \,cm$ $\frac{1}{v} =\frac{1}{-20}+\frac{1}{4} $ $\frac{1}{v} =\frac{-1+5}{20}=\frac{4}{20} $ $m =\frac{h_{i}}{h_{o}}=\frac{v}{u} $ $h_{i} =\frac{v h_{o}}{u} $ $=\frac{5 \times 2}{4}=\frac{10}{4}=2.5\, cm$