Science (086) — AI-generated practice question

Convex lens (f = +15 cm), object at u = –30 cm:

Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} = \frac{1}{15} + \frac{1}{(-30)} = \frac{2-1}{30} = \frac{1}{30}$

So v = +30 cm (positive, on opposite side of lens) → Image is real and inverted.

Reason: The object is placed at 2F₁ of the convex lens. A convex lens forms a real image when the object is placed beyond its focus.

Concave lens (f = –15 cm), object at u = –30 cm:

$\frac{1}{v} = \frac{1}{-15} + \frac{1}{-30}$...

Wait — $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-15} + \frac{1}{-30}$...

$\frac{1}{v} = -\frac{2}{30} - \frac{1}{30} = -\frac{3}{30}$...

Correcting: $\frac{1}{v} = \frac{1}{-15} - \frac{1}{-30} = -\frac{2}{30} + \frac{1}{30} = -\frac{1}{30}$

So v = –10 cm (negative, same side as object) → Image is virtual and erect.

Reason: A concave lens always forms a virtual, erect and diminished image, regardless of the position of the object.

Source: Chapter 9, Section 9.3.4 & 9.3.6

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• Convex lens: object at 30 cm = 2F, so image forms at 2F on the other side — real and inverted. Examiners want you to apply the lens formula and state the sign of v.
• Concave lens: always produces a virtual image (v is always negative). The lens formula confirms v = –10 cm. Quoting the textbook rule ("a concave lens always gives a virtual, erect, diminished image") scores the reason mark.
• Always use the New Cartesian Sign Convention: distances measured from optical centre; object distance u is negative; f of concave lens is negative.