Question

A concave lens forms the image of an object which is:

• A. Virtual, inverted and diminished
• B. Virtual, upright and diminished
• C. Virtual, inverted and enlarged
• D. Virtual, upright and enlarged

Hint:

Concave lenses always form virtual, upright, and diminished images, irrespective of the object's distance from the lens.

Answer 1

The Correct Option is B

When dealing with lenses, it's important to understand how they form images of objects placed at various positions relative to the lens. For a concave lens, which is also known as a diverging lens, the properties of the image formed are consistent:

• Virtual: The image formed by a concave lens is not actuable on a screen. Virtual images are formed on the same side of the lens as the object.
• Upright: The image retains the same orientation as the object, so it appears in the same upright direction.
• Diminished: The image is smaller than the object due to the lens's diverging properties.

Therefore, for a concave lens, the image of an object will be virtual, upright, and diminished.

Answer 2

A concave lens is a diverging lens, and its behaviour can be checked directly using the thin lens formula \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \), where distances are measured from the lens with the usual sign convention (object distance \( u \) negative, and focal length \( f \) negative for a concave lens). Let's use this to test each option instead of just quoting the standard property.

1. Virtual, inverted and diminished: An inverted image would need the magnification \( m = v/u \) to be negative, which needs \( v \) and \( u \) to have opposite signs. Since \( u \) is negative (the object is to the left of the lens) and solving the lens formula with a negative \( f \) always gives a negative \( v \) too, \( v \) and \( u \) share the same sign, so the image can never be inverted for a concave lens.
2. Virtual, upright and diminished: Solving \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \) with both \( f \) and \( u \) negative gives a negative \( v \) that is always smaller in size than \( u \), placing the image on the same side as the object (virtual) with \( |v| < |u| \) (diminished), and since \( v \) and \( u \) carry the same sign, the magnification \( m = v/u \) is positive (upright).
3. Virtual, inverted and enlarged: An enlarged image would need \( |v| > |u| \), but the lens equation for a concave lens always keeps \( |v| < |u| \) no matter where the object is placed, so a concave lens can never enlarge the image.
4. Virtual, upright and enlarged: This has the right sign behaviour for upright and virtual, but again fails on size, since \( |v| \) can never exceed \( |u| \) for a diverging lens.

Working through the lens formula with the correct sign convention rules out inversion and enlargement for any object position, leaving only the image being virtual, upright and diminished.

Therefore, the correct answer is Virtual, upright and diminished.