An object 2 m tall is placed at a distance of 50 cm in front of a convex lens. The lens has a power of +5 D. Calculate the height…

Question

An object 2 m tall is placed at a distance of 50 cm in front of a convex lens. The lens has a power of +5 D. Calculate the height and nature of the image formed by the lens.

1. A. 1 m, virtual and erect
2. B. 1.33 m, real and inverted (Correct answer)
3. C. 1.5 m, real and inverted
4. D. 1.6 m, virtual and erect

Correct Answer

Option B — 1.33 m, real and inverted

Detailed Solution & Explanation

The correct answer is 1.33 m, real and inverted.

Key Points

• Step 1 — find the focal length from the power.
  • $P = \frac{1}{f \text{ (in metres)}} \;\Rightarrow\; f = \frac{1}{5} = 0.2 \text{ m} = 20 \text{ cm}$
• Step 2 — apply the lens formula with $u = -50$ cm (object on the left) and $f = +20$ cm (convex lens):
  • $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
  • $\frac{1}{v} = \frac{1}{20} + \frac{1}{-50} = \frac{5 - 2}{100} = \frac{3}{100}$
  • $v = \frac{100}{3} \approx +33.33 \text{ cm}$
• Step 3 — find the magnification.
  • $m = \frac{v}{u} = \frac{100/3}{-50} = -\frac{2}{3}$
• Step 4 — find the image height.
  • $h' = m \times h = -\frac{2}{3} \times 2 = -1.33 \text{ m}$
• The positive $v$ means the image forms on the opposite side of the lens, so it is real; the negative height means it is inverted. Size 1.33 m.

Shortcut Trick

• The object sits at $u = 50$ cm with $f = 20$ cm, so it lies between $f$ and $2f$... in fact beyond $2f = 40$ cm. For a convex lens with the object beyond $2f$, the image is always real, inverted and diminished, forming between $f$ and $2f$. That alone eliminates both "virtual and erect" options without any calculation.

Additional Information

• Image formation by a convex lens:

• Sign convention: distances measured in the direction of incident light are positive; heights above the principal axis are positive. A real image always has negative magnification for a single lens, hence always inverted.
• Power is measured in dioptres (D); it is positive for converging (convex) lenses and negative for diverging (concave) lenses.