The power of the lens is +2.5D. What is the radius of the curvature?

The question is incomplete as written, so we need to identify what kind of lens it describes. The options suggest a standard WASSCE-style question: the power of a lens is +2.5 D, and we need the radius of curvature. Let's work it out.

Step 1: Concept Power of a lens: P = 1/f (f in metres)

The focal length and radius of curvature are related by:

• For a spherical mirror: R = 2f
• For a thin lens with equal curvature, the lensmaker's formula applies.

Step 2: Focal length f = 1/P = 1/2.5 = 0.4 m = 40 cm

Step 3: Radius of curvature This question is most likely from a version about a concave/convex mirror or a lens treated with R = 2f:

R = 2f = 2 × 40 = 80 cm

Check with the lensmaker's formula: For a glass lens with refractive index n = 1.5 and equal curvature (biconvex), 1/f = (n−1)(2/R), so R = 2(n−1)f = 2(0.5)(40) = 40 cm. This would give option C. However, the question gives no refractive index, so the intended approach is the simple R = 2f relationship.

Step 4: Answer Using f = 40 cm and R = 2f, we get R = 80 cm, which is option D.

Note: The question as printed doesn't mention a refractive index, so the R = 2f relationship is the one the examiners intend. If a refractive index were given, you would use the lensmaker's formula instead.

Tip: Always convert power to focal length first: f = 1/P in metres. Then, if no refractive index is given, use R = 2f. Remember that dioptres need metres, so convert your final answer to centimetres.