Question

In the figure shown , the radius of curvature of the left & right surfaces of the concave lens are $10\text{cm}$ & $\text{15}\text{cm}$ respectively. The radius of curvature of the mirror is $15\text{cm}$. Equivalent focal length of the combination is:

• A. The system behaves like a convex mirror of focal length $18\text{cm}$
• B. The system behaves like a concave mirror of focal length $18\text{cm}$
• C. The system behaves like a convex mirror of focal length $36\text{cm}$
• D. The system behaves like a concave mirror of focal length $36\text{cm}$

Answer 1

The correct option is B The system behaves like a concave mirror of focal length $18\text{cm}$


$\frac{1}{f_1}=(\frac{3}{2}-1)(\frac{-1}{10}-\frac{1}{15})=-\frac{1}{12};\frac{1}{f_2}=(\frac{4}{3}-1)\left(\frac{2}{15}\right)=\left(\frac{2}{45}\right);\frac{1}{f_m}=-\frac{2}{15}$

$\Rightarrow \frac{1}{f_l}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{-1}{12}+\frac{2}{45}=\frac{-7}{180}$

$\frac{1}{f_{eq}}=\frac{1}{f_m}-\frac{2}{f_l}=-\frac{1}{18}\Rightarrow f_{eq}=-18\text{cm}$

So, the combination behaves as a concave mirror