Question

A convex lens of refractive index $1.5$ has a focal length of $18cm$ in air. Calculate its focal length when it is immersed in water of refractive index $\frac{4}{3}.take{f}_1=18$.

• A. $f_1$
• B. $2f_1$
• C. $3f_1$
• D. $4f_1$

Answer 1

The correct option is D $4f_1$

At air, $\frac{1}{f_1}={(}_a{n}_g-1\left)\right(\frac{1}{R_1}-\frac{1}{R_2})$...(1)

At water, $\frac{1}{f}={(}_w{n}_g-1\left)\right(\frac{1}{R_1}-\frac{1}{R_2})$...(2)

$\frac{\left(1\right)}{\left(2\right)}\Rightarrow \frac{f}{f_1}=\frac{{}_a{n}_g-1}{{}_w{n}_g-1}$

$=\frac{1.5-1}{\frac{1.5}{4/3}-1}=\frac{0.5}{1/8}$

$f=4f_1$