Question

A thin convex lens of refractive index $1.5$ has $20cm$ focal length in air. If the lens is completely immersed in a liquid of refractive index $1.6,$ its focal length will be

• A. $-160cm$
• B. $-100cm$
• C. $+160cm$
• D. $+100cm$

Answer 1

Answer: (A)

$\frac{1}{f}=\left(a_a^{{\mu }_g}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\dots .\text{ (i) }$
Also, ${\mu }_g=\frac{{\mu }_g}{{\mu }_l}=\frac{1.5}{1.6}$
$\therefore \frac{1}{f^{\prime }}=\left({}_l{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{f^{\prime }}=\left(\frac{15}{16}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\dots \text{ (ii) }$
Dividing Equation (i) by Equation (ii), we get
$\frac{f}{f}=\frac{\left(\frac{15}{16}-1\right)}{(1.5-1)}=-\frac{1}{8}$
$f^{\prime }=-8f$
$=-160cm$