Question

A lens made from a material of refractive index 1.5 behaves as a converging lens in air. When placed in liquid of refractive index $\frac{8}{5}$, it will :

• A. still behave as a converging lens
• B. behave as a diverging lens
• C. have its focal length increased by 8 times
• D. have its focal length increased by 4 times

Answer 1

Answer: (B), (C)

The correct options are
B behave as a diverging lens
C have its focal length increased by 8 times

$\frac{1}{f_a}={(}_a{\mu }_g-1)[\frac{1}{R_1}-\frac{1}{R_2}]$
[$f_a$ = focal length of lens in air]
$\frac{1}{f_L}={(}_L{m}_g-1)[\frac{1}{R_1}-\frac{1}{R_2}]=\frac{{(}_L{\mu }_g-1)}{{}_a{\mu }_g-1}\frac{1}{f_a}$
$=\frac{(\frac{\frac{3}{2}}{\frac{8}{5}}-1)}{(\frac{3}{2}-1)}\frac{1}{f_a}=\frac{(\frac{15}{16}-1)}{\frac{1}{2}}\frac{1}{f_a}=-\frac{1}{8}.\frac{1}{f_a}$
$\therefore f_L=-8f_a$; negative sign indicate that lens is diverging.