Question

Two thin plano-concave lenses each made of glass of refractive index 1.5 have of curvature 10 cm. If space between lens is filled with oil of refractive index $\frac{8}{7}$ as shown in figure, then magnitude of focal length of the combination will be

• A. -14 cm
• B. 15 cm
• C. 18 cm
• D. 21 cm

Answer 1

Answer: (A)

-14 cm

$\text{we know,}$

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

$\begin{array}{c}\text{for plano concave lens}\\ n=1.5,-R_2=10cm,R_1=\infty \end{array}$

$\Rightarrow \frac{1}{f_1}=(1.5-1)(-\frac{1}{10})$

$\begin{array}{c}\Rightarrow f_1=-20cm\\ \text{for convex lens,}n=\frac{8}{7},R_1=\mathrm{10}cm=-R_2\end{array}$

$\begin{array}{c}\Rightarrow \frac{1}{f_2}=(\frac{8}{7}-1)(\frac{1}{10}+\frac{1}{10})\\ \Rightarrow \frac{1}{f_2}=\frac{1}{35}\Rightarrow f_2=35cm\end{array}$

$\begin{array}{c}\text{now, for the combination}\\ \begin{array}{c}1=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_1}\\ \frac{1}{F}=\frac{1}{-20}+\frac{1}{35}+\frac{1}{-20}\\ \frac{1}{F}=\frac{1}{35}-\frac{1}{10}=-\frac{25}{350}=\frac{-1}{14}\\ F=-14cm\end{array}\end{array}$