Question

A convex lens of focal length 15 cm, is forming a real image. If the size of image is same as the size of object, then position of object and position of image will be, respectively :

• A. \(-15 \, \text{cm and} -15 \, \text{cm from lens}\)
• B. \(-15 \, \text{cm and} +15 \, \text{cm from lens}\)
• C. \(-30 \, \text{cm and} +30 \, \text{cm from lens}\)
• D. \(-30 \, \text{cm and} -30 \, \text{cm from lens}\)

Hint:

For convex lens:

• Object at \(2f\) (\(u = -2f\)) \(\Rightarrow\) Image at \(2f\) (\(v = +2f\)), same size, real and inverted
• Object at \(f\) \(\Rightarrow\) Image at infinity
• Object between \(f\) and \(2f\) \(\Rightarrow\) Image beyond \(2f\), enlarged

Answer 1

The Correct Option is C

To solve this problem, let's use the lens formula and the magnification concept of a convex lens.

The focal length \(f\) of the lens is given as \(15 \, \text{cm}\).

The lens formula is:

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

where:

• \(f\) is the focal length,
• \(v\) is the image distance, and
• \(u\) is the object distance.

Since the size of the image is the same as the size of the object, the magnification \((m)\) is 1.

The magnification produced by a lens is also given by the formula:

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

Given \(m = 1\), we have:

\(\frac{v}{u} = 1 \Rightarrow v = u\)

Substitute \(v = u\) in the lens formula:

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

Thus:

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

Substituting \(f = 15 \, \text{cm}\), we get:

\(v = u = 2 \times 15 = 30 \, \text{cm}\)

For a real image formed by a convex lens, the object is placed on the opposite side of the lens than the image. Therefore, the object distance \(u\) is negative:

\(u = -30 \, \text{cm}\) and \(v = +30 \, \text{cm}\).

Hence, the position of the object and the position of the image are \(-30 \, \text{cm and} +30 \, \text{cm from the lens}\), respectively.

Therefore, the correct answer is: \(-30 \, \text{cm and} +30 \, \text{cm from lens}\).