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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 â€“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.