Question

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

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

Answer 1

Answer: (A)

$\frac{1}{f}=\left(a_{a}^{\mu_{g}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $
$=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \ldots . \text { (i) }$
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) $
$\frac{1}{f^{\prime}}=\left(\frac{15}{16}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \ldots \text { (ii) }$
Dividing Equation (i) by Equation (ii), we get
$\frac{f}{f}=\frac{\left(\frac{15}{16}-1\right)}{(1.5-1)}=-\frac{1}{8}$
$f^{\prime}=-8 f$
$=-160\, cm$