Question

A double convex lens of glass ($\mu =$1.5) is immersed in water($\mu =$4/3). If the focal length in air is $F$, focal length in water is

• A. $F$
• B. $F/2$
• C. $2F$
• D. $4F$

Answer 1

Answer: (D)

$4F$

$\frac{1}{F}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{F^1}=(\frac{\frac{3}{2}}{4/3}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{\frac{1}{F}}{\frac{1}{f^1}}=\frac{\frac{1}{2}}{\frac{1}{8}}$

$\Rightarrow f^1=4F$