The near point of a hypermetropic person is 80 cm from the eye. The power of lens, required to enable the person to read clearly a book held at 25 cm from the eye is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.75 D
A hypermetropic eye needs a converging (convex) lens that forms a virtual image of the book, held at 25 cm, at the near point of the eye, 80 cm away on the same side as the book. Use the lens formula with the sign convention (distances measured from the lens, positive along the incident light): \( u = -25\,\mathrm{cm} \), \( v = -80\,\mathrm{cm} \).
\( \displaystyle \frac{1}{f} = \frac{1}{v}-\frac{1}{u} = -\frac{1}{80}+\frac{1}{25} = \frac{-5+16}{400} = \frac{11}{400}\,\mathrm{cm^{-1}} \)
\( \displaystyle f = \frac{400}{11}\,\mathrm{cm} \approx 36.4\,\mathrm{cm} = 0.364\,\mathrm{m} \)
The power of a lens is \( P = 1/f \) with f in metres:
\( \displaystyle P = \frac{1}{0.364}\,\mathrm{D} = +2.75\,\mathrm{D} \)
Shortcut: The power needed is \( P = \frac{1}{0.25\,\mathrm{m}} - \frac{1}{0.80\,\mathrm{m}} = 4\,\mathrm{D} - 1.25\,\mathrm{D} = 2.75\,\mathrm{D} \).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
The Eye – “… Example 9.12 (a) The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?”