The power of a lens is +1.5D. Name the type of defects vision that can be corrected by using this lens. Find the focal length of the lens.
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The defect of vision is hypermetropia. It can be corrected by using convex lens. The focal length of lens will be:$\text{P}=\frac{1}{\text{f}}$
$\text{f} =\frac{1}{\text{P}} $ (in meter)
$\text{f}=\frac{1}{1.5\text{m}}$
$\text{f}=0.66\text{m}$
$\text{f} =\frac{1}{\text{P}} $ (in meter)
$\text{f}=\frac{1}{1.5\text{m}}$
$\text{f}=0.66\text{m}$
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