Question

A thin equiconvex lens is made of glass of R.I 1.5 and its focal length is 0.2 If it acts a concave lens of 0.5 m focal length when dipped in a liquid, the R.I of liquid is

• A. $\frac{17}{8}$
• B. $\frac{15}{8}$
• C. $\frac{13}{8}$
• D. $\frac{9}{8}$

Answer 1

The correct option is B $\frac{15}{8}$

$-\frac{f_1}{f_a}=\frac{({\mu }_5-1)}{(\frac{{\mu }_s}{{\mu }_l}-1)}-2.5=\frac{0.5}{(\frac{1.5}{{\mu }_l}-1)}-\frac{0.5}{0.2}=\frac{1.5-1}{(\frac{1.5}{{\mu }_1}-1)}\frac{1.5}{{\mu }_l}-1=-\frac{1}{5}\frac{1.5}{{\mu }_l}=1-\frac{1}{5}=\frac{4}{5}{\mu }_l=\frac{7}{4}=\frac{75}{40}=\frac{15}{8}$