Question
A thin convex lens made from crown glass $$\left( {\mu = \frac{3}{2}} \right)$$ has focal length $$f.$$ When it is measured in two different liquids having refractive indices $$\frac{4}{3}$$ and $$\frac{5}{3}$$ it has the focal lengths $${f_1}$$ and $${f_1}$$ respectively. The correct relation between the focal lengths is:
A.
$${f_1} = {f_2} < f$$
B.
$${f_1} < f$$ and $${f_2}$$ becomes negative
C.
$${f_2} < f$$ and $${f_1}$$ becomes negative
D.
$${f_1}$$ and $${f_2}$$ both become negative
Answer :
$${f_1} < f$$ and $${f_2}$$ becomes negative
Solution :
By Lens maker's formula for convex lens
$$\eqalign{
& \frac{1}{f} = \left( {\frac{\mu }{{{\mu _L}}} - 1} \right)\left( {\frac{2}{R}} \right) \cr
& {\text{for, }}{\mu _{{L_1}}} = \frac{4}{3},{f_1} = 4R \cr
& {\text{for, }}{\mu _{{L_2}}} = \frac{5}{3},{f_2} = - 5R \cr
& \Rightarrow \,\,{f_2} = \left( - \right)ve \cr} $$