Question

In the figure shown the radius of curvature of the left and right surface of the concave lens are 10 cm and 15 cm respectively. The radius of curvature of the mirror is 15 cm equivalent focal length of the combination is

• A. the system behaves like a convex mirror of focal length 18 cm
• B. the system behaves like a concave mirror of focal length 18 cm
• C. the system behaves like a convex mirror of focal length 36 cm
• D. the system behaves like a concave mirror of focal length 36 cm

Answer 1

Answer: (A)

the system behaves like a convex mirror of focal length 18 cm

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\begin{array}{c}\text{focal length of glass lens}=f_G\\ \frac{1}{f_G}=(\frac{3}{2}-1)[\frac{1}{-10}-\frac{1}{15}]\\ \frac{1}{f_G}=-\frac{1}{2}\left[\frac{3+2}{30}\right]=-\frac{1}{12}\\ \text{focal length of water lens}=f\omega \\ \frac{1}{f_{\omega }}=(\frac{4}{3}-1)[\frac{1}{1.5}-\frac{1}{-15}]=\frac{1}{3}\times \frac{2}{15}=\frac{2}{45}\end{array}$

focal length of mirror $=\frac{\text{Radius of curvature}}{2}$

$f_m=\frac{-15}{2}$

for combination

$\frac{1}{f_{\text{resultant}}}=\frac{1}{f_{\text{mirror}}}-\frac{2}{f_{\text{lens}}}$

So, Here $\frac{1}{f}=\frac{1}{f_m}-2[\frac{1}{f_G}+\frac{1}{f_{\omega }}]$

$\frac{1}{f}=\frac{-1}{\frac{15}{2}}-2[\frac{-1}{12}+\frac{2}{45}]$

$\frac{1}{f}=\frac{-2}{15}+\frac{1}{6}-\frac{4}{45}$

$\begin{array}{cc}\frac{1}{f}=\frac{1}{6}-\frac{60}{45}& =\frac{1}{6}-\frac{2}{9}\\ \frac{1}{f}& =\frac{3-2}{18}\\ f^{}=+18cm& \end{array}]\to$ convex mirror

So option A is correct