To obtain a virtual, erect and magnified image from a thin converging lens, an object must be placed

Analyze the question: We need the object position in front of a converging (convex) lens that gives an image that is virtual, erect and magnified.

Concept: The image formed by a converging lens depends on where the object is placed relative to the focal length (f). Use the lens formula:

1/f = 1/u + 1/v

and magnification m = v/u.

Going through each position:

| Object position | Image nature | |---|---| | Beyond 2f | Real, inverted, diminished | | At 2f | Real, inverted, same size | | Between f and 2f | Real, inverted, magnified | | At f | Image at infinity | | Between f and the lens (u < f) | Virtual, erect, magnified |

Check with the formula: Take f = 10 cm and u = 6 cm (object between the lens and the focus).

1/v = 1/f − 1/u = 1/10 − 1/6 = (3 − 5)/30 = −2/30

v = −15 cm

• The negative sign means the image is virtual (on the same side as the object).
• m = v/u = 15/6 = 2.5, so the image is magnified and erect.

This is how a magnifying glass works: you hold the object closer to the lens than its focal length.

Evaluating the options:

• A. Between the focus and the lens: correct.
• B. At the focus: the image forms at infinity, so it is not a usable virtual image.
• C. Between f and 2f: the image is real and inverted.
• D. Beyond 2f: the image is real, inverted and diminished.

Answer: A

Exam tip: Remember "a magnifying glass needs the object inside the focus." A virtual image always means the object is within f for a converging lens. Real images are always inverted, and virtual ones are always erect.