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 $10\, cm$ 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 $f^{'}$ then
$\frac{1}{f}=\frac{1}{f^{'}}+\frac{1}{f^{'}} $
i.e., $ f^{'}=2 f$
i.e., each part will have focal length $2 f$.
Now if these parts are put in contact as in figure $(2)$,
then resultant focal length of the combination will be
$\frac{1}{F}=\frac{1}{2 f}+\frac{1}{2 f}$
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
$\frac{1}{F^{'}} =\left({ }_{l} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\,\,\,...(ii)$
$\therefore \frac{F^{'}}{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 $\frac{F^{'}}{f}=\frac{0.5}{\left(\frac{9}{8}-1\right)}=0.5 \times 8$
$\therefore F^{'}=0.5 \times 8 \times 10=40 \,cm$