Understanding Image Formation by a Convex Lens

The question asks us to determine the nature, position, and size of the image formed by a convex lens when an object is placed at a specific distance from it. We are given that the object is placed at a distance of \(\frac{f}{2}\) from a convex lens, where \(f\) is the focal length of the lens.

Applying the Lens Formula

To find the characteristics of the image, we can use the lens formula, which relates the object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)) of a lens:

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

We need to use the proper sign conventions. For a convex lens, the focal length \(f\) is considered positive. The object is placed to the left of the lens, so the object distance \(u\) is negative. Given that the object distance is \(\frac{f}{2}\), we have:

\(u = -\frac{f}{2}\)

\(f = +f\) (for a convex lens)

Now, we substitute these values into the lens formula:

\(\frac{1}{v} - \frac{1}{\left(-\frac{f}{2}\right)} = \frac{1}{f}\)

\(\frac{1}{v} + \frac{2}{f} = \frac{1}{f}\)

To find the image distance \(v\), we rearrange the equation:

\(\frac{1}{v} = \frac{1}{f} - \frac{2}{f}\)

\(\frac{1}{v} = \frac{1-2}{f}\)

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

\(v = -f\)

Interpreting the Image Distance

The image distance \(v = -f\). The negative sign for \(v\) indicates that the image is formed on the same side of the lens as the object (to the left of the lens). The distance of the image from the lens is \(f\). This position is the first principal focus (\(F_1\)) of the convex lens, located on the object side.

Since the image distance \(v\) is negative, the image formed is a virtual image.

Calculating Magnification

To determine the size and orientation of the image, we calculate the magnification \(m\), which is given by the formula:

\(m = \frac{v}{u}\)

Substitute the values of \(v\) and \(u\):

\(m = \frac{-f}{-\frac{f}{2}}\)

\(m = \frac{f}{\frac{f}{2}}\)

\(m = f \times \frac{2}{f}\)

\(m = 2\)

Interpreting Magnification

The magnification \(m = 2\). The positive sign of \(m\) indicates that the image is erect (same orientation as the object). The magnitude of \(m\) (\(|m|=2\)) indicates that the image is magnified to double the size of the object.

Summary of Image Properties

• Position: \(v = -f\) (at a distance \(f\) on the same side as the object, i.e., at the first principal focus \(F_1\)).
• Nature: Virtual and erect (since \(v\) is negative and \(m\) is positive).
• Size: Magnified, double the size of the object (since \(|m|=2\)).

Comparing with Options

Let's check our findings against the given options:

1. At one of the foci, virtual and double of its size: This matches our results. The image is formed at a distance \(f\) on the object side (one of the foci), it is virtual, and its size is double that of the object.
2. At \(\frac{3}{2}\) f, real and inverted: Our image is virtual, not real, and at distance \(f\), not \(\frac{3}{2}f\).
3. At 2f, virtual and erect: Our image is at distance \(f\), not 2f.
4. None of these: Option 1 matches our findings.

When an object is placed between the optical center and the principal focus (\(O\) and \(F_1\)) of a convex lens, a virtual, erect, and magnified image is formed on the same side as the object. Our object distance \(u = f/2\) falls within this range (since \(u < f\)), confirming the expected virtual image.