Question

The two surfaces of a biconvex lens has same radii of curvatures. This lens is made of glass of refractive index $1.5$ and has a focal length $10cm$ in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. The two pieces are glued such that the convex surfaces touch each other. If this combination lens is immersed in water
(refractive index $=$ $4/3$ ), its focal length (in $cm$ ) is :

• A. 5
• B. 10
• C. 20
• D. 40

Answer 1

Answer: (D)

If $a$ lens of focal length $f$ is divided into two equal parts as shown in figure $(1)$ and each has a focal length $f^{{}^{\prime }}$ then
$\frac{1}{f}=\frac{1}{f^{{}^{\prime }}}+\frac{1}{f^{{}^{\prime }}}$
i.e., $f^{{}^{\prime }}=2f$
i.e., each part will have focal length $2f$.
Now if these parts are put in contact as in figure $(2)$,
then resultant focal length of the combination will be
$\frac{1}{F}=\frac{1}{2f}+\frac{1}{2f}$
i.e., $F=f$ (initial value)

For this combination,
$\frac{1}{F}=\left({}_a{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)...(i)$
Now, if this combination is immersed in liquid, then
$\frac{1}{F^{{}^{\prime }}}=\left({}_l{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)...(ii)$
$\therefore \frac{F^{{}^{\prime }}}{f}=\frac{\left(a{\mu }_g-1\right)}{\left({\mu }_g-1\right)}=\frac{(1.5-1)}{\left(\frac{\frac{3}{2}}{\frac{4}{4}}-1\right)}$
or $\frac{F^{{}^{\prime }}}{f}=\frac{0.5}{\left(\frac{9}{8}-1\right)}=0.5\times 8$
$\therefore F^{{}^{\prime }}=0.5\times 8\times 10=40cm$