Make a diagram to show how hypermetropia is corrected.

Question.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is 25 cm.

solution:

Here, y = 1 m

$\frac{1}{f}=\frac{1}{0.25}-\frac{1}{y}=\frac{1}{0.25}-\frac{1}{1}$

$=\frac{100}{25}-1=4-1=+3$

or $\mathrm{f}=+(1 / 3)=+0.3333 \mathrm{~m}=+33.33 \mathrm{~cm}$

Power, $P=\frac{1}{f}=+3$ dioptres.

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is 25 cm.

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