Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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 $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App