Many optical instruments consist of a number of lenses. They are combined to increase the magnification and sharpness of the image. The net power (P) of the lenses placed in contact is given by the algebraic sum of the powers of the individual lenses P1, P2, P3 …
P = P1 + P2 + P3 …
This is also termed as the simple additive property of the power of lens, widely used to design lens systems of cameras, microscopes and telescopes. These lens systems can have a combination of convex lenses and also concave lenses.
Read the case carefully and answer the questions that follow.
Generated by claude-sonnet-4-6 · 2026-06-15 06:50 · grounding stimulus
Model Answer
(a) Net Power = +4 D + (−2 D) = +2 D
Since the net power is positive, the combination is convergent.
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(b) Power P = −2.5 D
$$f = \frac{1}{P} = \frac{1}{-2.5} = -0.4 \text{ m} = -40 \text{ cm}$$
The focal length is −40 cm (concave lens).
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(c) Power = +0.1 D → $f = \frac{1}{0.1} = 10$ m = 1000 cm
Object distance = 20 cm. Since the object is placed between the optical centre and focus (20 cm < 1000 cm = f), the image formed is:
- Virtual, erect, and magnified
- Formed on the same side as the object
(Ray diagram: Two rays from object tip — one parallel to principal axis refracting through F, one through optical centre — diverging rays extended back to meet behind the lens on the same side.)
Source: Light – Reflection and Refraction, Power of a Lens
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Explanation
- (a) Simply add the powers algebraically; positive result → converging.
- (b) Use $f(\text{m}) = 1/P$; negative f confirms concave lens.
- (c) First find f from P. Since object distance (20 cm) < f (1000 cm), it's the "between O and F" case → virtual, erect, magnified image. Examiners award 1 mark for correct ray diagram and 1 mark for stating nature/position of image. Draw and label clearly.