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(a) A person is suffering from both myopia and hypermetropia. (i) What kind of lenses can correct this defect? (ii) How are these lenses prepared?(b) A person needs a lens of power + 3D for correcting his near vision and –3D for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects.
(a) If a person is suffering from both myopia and hypermetropia.
(i) Bifocal lenses can correct this defect.
(ii) These lenses are prepared by combining both the lenses in spectacles, such that its upper part of the spectacles consists of a concave lens (to correct myopia) and the lower part consists of a convex lens (to correct hypermetropia).
(b) Given:
Power, $P_1$ = $+$3D (for correcting near vision)
Power, $P_2$ = $-$3D (for correcting distant vision)
To find: Focal length of the lenses, $f$.
Solution:
We know that power of lens is given as-
$P=\frac {1}{f}$
Substituting the given values we get-
$P_1=\frac {1}{f_1}\Rightarrow3=\frac {1}{f}\Rightarrow f_1=\frac {1}{3}\Rightarrow f_1=0.33m\Rightarrow+33.3cm$
$P_2=\frac {1}{f_2}\Rightarrow-3=\frac {1}{f}\Rightarrow f_2=-\frac {1}{3}\Rightarrow f_2=-0.33m\Rightarrow-33.3cm$
Therefore, the focal lengths of the lenses required to correct these defects are +33.3 cm and -33.3 cm respectively.
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