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.
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.
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.