Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
Generated by claude-sonnet-4-6 · 2026-06-15 06:47 · grounding stimulus
Model Answer
(a)
Using lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Given: f = +20 cm, u = −40 cm
$\frac{1}{v} = \frac{1}{20} + \frac{1}{(-40)} = \frac{2-1}{40} = \frac{1}{40}$
∴ v = +40 cm
Image is formed at 40 cm on the other side of the lens; it is real, inverted, and same size as the object.
---
(b)
```
F' O F
←--------|---------|--------|--------→
\ | /
\ virtual, /
\ erect, /
\diminish /
\ image /
```
When object is between O and F of a concave lens, the image is formed on the same side as the object — virtual, erect, and diminished.
(Ray diagram: two rays diverge after refraction; their extensions meet on the same side as the object.)
---
(c)
Given: f₁ = +30 cm, f₂ = −15 cm
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \frac{1}{-15} = \frac{1-2}{30} = \frac{-1}{30}$$
∴ F = −30 cm
Power: $P = \frac{1}{F(\text{in m})} = \frac{1}{-0.30}$
∴ P = −3.33 D
The combination acts as a concave (diverging) lens.
Source: Light – Reflection and Refraction, Lens Formula and Power of a Lens
---
Explanation
- (a) Apply the standard lens formula with sign convention (u is negative). At 2f, image forms at 2f — real, inverted, same size. Examiners expect the formula, substitution, and nature of image.
- (b) A ray diagram must show at least two rays and indicate the virtual, erect, diminished image on the same side as the object. Even a rough labeled sketch earns the mark.
- (c) Use $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ and $P = P_1 + P_2$. State the sign and nature of the combination — concave — to get full marks. Remember to convert cm to metres for power.