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, $\frac{f_m}{f}=\frac{(\mu -1)}{\left(\frac{\mu }{{\mu }_m}-1\right)}$
$<br/>\Rightarrow \frac{\left(\frac{3}{2}-1\right)}{\left(\frac{3/2}{4/3}-1\right)}=4<br/>$
$<br/>\Rightarrow f_1=4f<br/>$
$<br/>\frac{f_2}{f}=\frac{\left(\frac{3}{2}-1\right)}{\left(\frac{3/2}{5/3}-1\right)}=-5<br/>$
$<br/>\Rightarrow f_2<0<br/>$