NCERT · Class 10 · Science · The Human Eye and Colourful WorldThe near point of a hypermetropic person is 1 m. What is the power of the lens required to read a book clearly kept at 25 cm?

Given:

• Near point of the hypermetropic person (image distance, $v$) = $-1\text{ m} = -100\text{ cm}$
• Normal near point for clear vision (object distance, $u$) = $-25\text{ cm}$
• Focal length of the lens ($f$) = ? \nUsing the lens formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$| \nSubstitute the values into the equation: $$\frac{1}{f} = \frac{1}{-100} - \frac{1}{-25}$| $$\frac{1}{f} = -\frac{1}{100} + \frac{1}{25}$|\nTaking the LCM ($100$): $$\frac{1}{f} = \frac{-1 + 4}{100} = \frac{3}{100}$| $$f = \frac{100}{3}\text{ cm} = \frac{1}{3}\text{ m} = +0.33\text{ m}$| \nNow, calculate the power of the lens ($P$): $$P = \frac{1}{f\text{ (in meters)}} = \frac{1}{1/3} = +3.0\text{ D}$| \nConclusion:\nThe person requires a convex lens of focal length $+33.3\text{ cm}$ and a power of $+3.0\text{ D}$ to correct this hypermetropia.