The two surfaces of a biconvex lens has same radii of curvatures. This lens is made of glass of refractive…
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 $10 \mathrm{~cm}$ 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 $\mathrm{cm}$ ) is :
5
10
20
40
Solution
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
$\frac{1}{f}=\frac{1}{f^{\prime}}+\frac{1}{f^{\prime}} \quad \text { i.e., } f^{\prime}=2 f$
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 '
$\frac{1}{F}=\frac{1}{2 f}+\frac{1}{2 f} \text { i.e., } F=f \quad \text { (initial value) }$
For this combination,
$\frac{1}{F}=\left({ }_a \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ $\ldots$ (i)
Now, if this combination is immersed in liquid, then
$\frac{1}{F^{\prime}}=\left({ }_l \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ $\ldots$ (ii)
$\therefore \frac{F^{\prime}}{f}=\frac{\left({ }_a \mu_g-1\right)}{\left.{ }_l \mu_g-1\right)}=\frac{(1.5-1)}{\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)}$
or $\frac{F^{\prime}}{f}=\frac{0.5}{\left(\frac{9}{8}-1\right)}=0.5 \times 8$
$\therefore F^{\prime}=0.5 \times 8 \times 10=40 \mathrm{~cm}$
