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 $f_{2}$ respectively. The correct relation between the focal lengths is

• A. $f_1 - f_2 <\, f$
• B. $f_2 >\, f$ and $f_2$ becomes negative
• C. $f_2 >\, f$ and $f_1$ becomes negative
• D. $f_1 $ and $f_2 $ both become negative

Answer 1

Answer: (B)

Using lens makers formula and taking ratios, $\frac{ f _{ m }}{ f }=\frac{(\mu-1)}{\left(\frac{\mu}{\mu_{ m }}-1\right)}$
$
\Rightarrow \frac{\left(\frac{3}{2}-1\right)}{\left(\frac{3 / 2}{4 / 3}-1\right)}=4
$
$
\Rightarrow f _{1}=4 f
$
$
\frac{f_{2}}{f}=\frac{\left(\frac{3}{2}-1\right)}{\left(\frac{3 / 2}{5 / 3}-1\right)}=-5
$
$
\Rightarrow f _{2}<0
$