Question

the radius of curvature of eacjhface of biconcave lens made of glass of referactive index 1.5 is 30 cm. calculate the focal length of the lens in air

Answer 1

The relation between focal length in air and the radius of curvature is given by $\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, here R₁=30cm and R₂=-30cm, and n=1.5. Using this values in the above equation we get, $$\begin{gathered} \frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{30}-\left(-\frac{1}{30}\right)\right)=\frac{1}{30} \\ \Rightarrow f=30\text{cm} \end{gathered}$$