Question

A lens made of glass of refractive index $\frac{3}{2}$ has focal length f. Find its focal length when it is immersed in water of refractive index $\frac{4}{3}$.

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

Answer 1

The correct option is D $4f$

From lens makers formula

$\frac{1}{f}=({\mu }_{ga}-1)(\frac{1}{R_1}-\frac{1}{R_2}).....\left(1\right)\frac{1}{f^{\prime }}=({\mu }_{gw}-1)(\frac{1}{R_1}-\frac{1}{R_2}).....\left(2\right)\text{On Dividing (1) by (2)}\frac{f^{\prime }}{f}=\frac{({\mu }_{ga}-1)}{({\mu }_{gw}-1)}=\frac{(\frac{{\mu }_g}{{\mu }_a}-1)}{(\frac{{\mu }_g}{{\mu }_w}-1)}\frac{f^{\prime }}{f}=\frac{(\frac{3}{2}-1)}{(\frac{\frac{3}{2}}{\frac{4}{3}}-1)}=\frac{\left(\frac{3-2}{2}\right)}{(\frac{9}{8}-1)}\frac{f^{\prime }}{f}=\frac{\frac{1}{2}}{\frac{9-8}{8}}=\frac{\frac{1}{2}}{\frac{1}{8}}\frac{f^{\prime }}{f}=\frac{1}{2}\times \frac{8}{1}\frac{f^{\prime }}{f}=\frac{1}{2}\times \frac{8}{1}{f}^{\prime }=4f$