A \( 20 \) D lens is used as a magnifier. Where should the object be placed to obtain maximum angular magnification for a normal eye (near point \( = 25 \) cm)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -4.16 \) cm
A simple magnifier gives its maximum angular magnification when the final image is formed at the near point, D = 25 cm. Using the Cartesian sign convention (distances measured from the lens, positive in the direction of the incident light):
\( \displaystyle P = 20\,\mathrm{D} \ \Rightarrow\ f = \frac{1}{20}\,\mathrm{m} = 5\,\mathrm{cm} \)
\( v = -25\,\mathrm{cm} \quad \text{(virtual image on the same side as the object)} \)
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \ \Rightarrow\ \frac{1}{u} = \frac{1}{v} - \frac{1}{f} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25} \)
\( \displaystyle u = -\frac{25}{6}\,\mathrm{cm} \approx -4.17\,\mathrm{cm} \)
The paper gives this value as \( -4.16 \) cm (25/6 = 4.166..., shortened to two decimal places). The negative sign shows that the object is on the incident-light side of the lens, about 4.2 cm from it.
Taking \( v = +25 \) cm by mistake (a real image on the far side of the lens) gives \( 1/u = 1/25 - 1/5 = -4/25 \), that is \( u = -6.25 \) cm, the value in Options (3) and (4); but the image formed by a magnifier is virtual, so v is negative.
Note: The question asks where the object should be placed and does not say whether a signed value or a distance is wanted. If only the distance from the lens is meant, Option (1), 4.16 cm, gives the same position without the sign; the signed value of Option (2) is the complete answer in the Cartesian convention.