Question

For the shown system, the radius of curvature $\left(R\right)$ of the concave surface of the lens $\text{Q}$ is double the radius of curvature of the convex surface. If the plane surface of $\text{P}$ is silvered as shown in figure, the system acts as

• A. convex mirror of focal length $24\text{cm}$
• B. concave mirror of focal length $8\text{cm}$
• C. concave mirror of focal length $24\text{cm}$
• D. convex mirror of focal length $8\text{cm}$

Answer 1

The correct option is C concave mirror of focal length $24\text{cm}$

Combined focal length of lenses,

$\frac{1}{f_l}=({\mu }_Q-1)(\frac{1}{R_{1Q}}-\frac{1}{R_{2Q}})+({\mu }_P-1)(\frac{1}{R_{1P}}-\frac{1}{R_{2P}}).......\left(1\right)$

Here,
${\mu }_P=1.8=$refractive index of lens $\text{P}$ with respect to medium,

${\mu }_Q=1.2=$refractive index of lens $\text{Q}$ with respect to medium,

$R_{1Q}=$Radius of concave surface$=+24\text{cm}$,

$R_{2Q}=$radius of convex surface$=+48\text{cm}$,

$R_{1P}=$radius of convex surface$=+48\text{cm}$,

$R_{2P}=$radius of plane surface$=\infty$,

Substituting the values in $\left(1\right)$,

$\Rightarrow \frac{1}{f_l}=(1.2-1)(\frac{1}{(+24)}-\frac{1}{(+48)})+(1.8-1)(\frac{1}{(+48)}-\frac{1}{\infty })$

$\Rightarrow \frac{1}{f_l}=\frac{1}{48}......\left(2\right)$

Now, combined power of system,
$P=P_M-2P_L$

$\therefore \frac{1}{F}=\frac{1}{f_m}-\frac{2}{f_l}$

where,
$f_m=$ focal length of plane mirror$=\infty$

$\Rightarrow \frac{1}{F}=\frac{1}{\infty }-2\times \frac{1}{48}$

$\Rightarrow F=-24\text{cm}$

Since, the combined focal length is negative so, the combination will act as concave mirror of focal length $24\text{cm}$.