Question

A plano-convex lens $\text{P}$ and a concavo-convex lens $\text{Q}$ are in contact as shown in figure. The refractive index of the material of the lens $\text{P}$ and $\text{Q}$ are $1.8$ and $1.2$ respectively. The radius of curvature of the concave surface of the lens $\text{Q}$ is double the radius of curvature of the concave surface. The convex surface of $\text{Q}$ is silvered. The radius of curvature of common surface if focal length of the system is $-8\text{cm}$ will be

• A. $48\text{cm}$
• B. $24\text{cm}$
• C. $12\text{cm}$
• D. $8\text{cm}$

Answer 1

The correct option is A $48\text{cm}$

Given,
refractive index of the material $\text{P}$, ${\mu }_P=1.8$
refractive index of the material $\text{Q}$, ${\mu }_Q=1.2$
focal length of the system, $f=-8\text{cm}$

Formula for the focal length of the combination of the thin lenses,

$\frac{1}{F}=\frac{1}{F_P}+\frac{1}{F_Q}....\left(1\right)$

From lens makers formula plano-convex lens $\text{P}$,

$\frac{1}{F_P}=({\mu }_P-1)(\frac{1}{R_1}-\frac{1}{R_2})....\left(2\right)$

where,
$R_1=$ Radius of curvature of plane surface$=\infty$
$R_2=$ Radius of curvature of convex surface$=-2R$

Substituting the values in $2$,

$\frac{1}{F_P}=(1.8-1)(\frac{1}{\infty }-\frac{1}{-2R})$

$\therefore \frac{1}{F_P}=\frac{0.4}{R}.....\left(3\right)$

Similarly applying the lens makers formula for concavo-convex lens $\text{Q}$,

$\frac{1}{F_Q}=({\mu }_Q-1)(\frac{1}{R_1}-\frac{1}{R_2})....\left(4\right)$

where,
$R_1=$ Radius of curvature of convex surface$=-2R$
$R_2=$ Radius of curvature of concave surface$=-R$

Substituting the values in $4$,

$\frac{1}{F_Q}=(1.2-1)(\frac{1}{-2R}-\frac{1}{-R})$

$\therefore \frac{1}{F_Q}=\frac{0.1}{R}.....\left(5\right)$

Now, from $\left(1\right)$, $\left(3\right)$ and $\left(5\right)$,

$\frac{1}{F}=\frac{0.4}{R}+\frac{0.1}{R}=\frac{0.5}{R}$

Now, combined power of system,
$P=P_M-2P_L......\left(6\right)$

where,
$P_M=$ Power of concave mirror$=\frac{-2}{R}$
$P_L=$ Power of combination of lens$=\frac{1}{F}=\frac{0.5}{R}$

Substituting the values in $\left(6\right)$,

$\Rightarrow \frac{-1}{8}=\frac{2}{-R}-2\times \frac{0.5}{R}$

$\Rightarrow \frac{3}{R}=\frac{1}{8}$

$\Rightarrow R=24\text{cm}$

Therefore, radius of curvature of common surface is
$2R=2\times 24=48\text{cm}$

Hence, correct option is (a).