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A lens has a power of +5 diopters in air. What will be power if completely immersed in water? $_a \mu_w =\large\frac{4}{3}$ and $_a \mu_g =\large\frac{3}{2}$

$(a)\;1.25\;diopter \\ (b)\;2 diopter \\ (c)\;2.25\;diopter \\ (d)\;1\;diopter $

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Let $f_a$ and $f_m$ be focal length of lens in air and water respectively , then
$P_a= \large\frac{1}{f_a}$
or $+5 =\large\frac{1}{f_a}$
$f_a =0.2 \quad m=20\;cm$
Now $\large\frac{1}{f_a}=(_a \mu _g-1) \bigg[ \large\frac{1}{R_1}-\frac{1}{R_2} \bigg]$
and $\large\frac{1}{f_w}=(_w \mu _g-1) \bigg[ \large\frac{1}{R_1}-\frac{1}{R_2} \bigg]$
Divide equations we get,
$\large\frac{f_w}{f_a} =\bigg[ \large\frac{a \mu _g-1}{w \mu _g -1}\bigg]$
Again $_w \mu _g=\large\frac{_a \mu _g}{_a \mu _w}$
$\qquad= \large\frac{\Large\frac{3}{2}}{\Large\frac{4}{3}}$
$\qquad =\large\frac{9}{8}$
$\therefore \large\frac{f_w}{f_a} = \large\frac{\Large\frac{3}{2}-1}{\Large\frac{9}{8} -1}$
$\qquad= \large\frac{\Large\frac{1}{2}}{\Large\frac{1}{8}}$$=4$
$f_w=f_a \times 4$
$\qquad = 20 \times 4 $
$\qquad=0.8 mm$
$P_w =\large\frac{1}{f_w}=\frac{1}{0.8}$
$\qquad= 1.25 $ diopters
Hence a is the correct answer.
answered Feb 12, 2014 by meena.p

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