Question

A thin lens of focal length $+12cm$ is immersed in water $(\mu =1.33)$ its new focal length is $({\mu }_8=\frac{3}{2})$

• A. $24cm$
• B. $36cm$
• C. $47cm$
• D. $12cm$

Answer 1

The correct option is C $47cm$

Lens makers formula for in air:

$\frac{1}{12}=\left(\frac{{\mu }_g}{{\mu }_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Lens makers formula for in water:

$\frac{1}{f}=\left(\frac{{\mu }_g}{{\mu }_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Dividing the two equations we get,

$\frac{f}{12}=\frac{\left(\frac{{\mu }_g}{{\mu }_a}-1\right)}{\left(\frac{{\mu }_g}{{\mu }_w}-1\right)}$

$\frac{f}{12}=\frac{\left(1.5-1\right)}{\left(\frac{1.5}{1.33}-1\right)}$

$f=\frac{0.5}{0.12}\times 12=46.94$

$\therefore f\simeq 47cm$

Hence the correct answer is option (C).