Question

A biconvex lens has radii of curvature $20cm$ each.if the refractive index of the material of the lens is$1.5$.Its Focal length is?

• A. $20$
• B. $1.5$
• C. $7.89$
• D. $8.2$

Answer 1

Answer: (A)

$20$

$\begin{array}{c}{R}_1=20cm\\ R_2=-20cm\\ \mu =1.5\text{[refractive index]}\end{array}$

$\begin{array}{c}\text{Using formula:}\\ \begin{array}{ccc}\frac{1}{f}& =[\frac{{\mu }_1}{{\mu }_2}-4][\frac{1}{R_1}-\frac{1}{R_2}]& \\ [{\mu }_1=1.5;& {\mu }_2=1& \text{as its in air}]\\ \frac{y}{f}=[\frac{1.5}{1}-4][\frac{1}{20}-\frac{1}{(-20)}]& \Rightarrow \left[0.5\right][\frac{1}{20}+\frac{1}{20}]\Rightarrow \left[0.5\right]\left[\frac{2}{20}\right]& \\ \frac{1}{f}& =\frac{4}{20};f=20cm& \end{array}\end{array}$

$\begin{array}{c}\text{Remember it as a result that for a biconvese}\\ \text{lens for}\left|R_1\right|=\left|R_2\right|\text{and Rept in air made of}g\text{lass}(\mu =1.5)\\ f\text{is always equal to}\left|R_1\right|\text{or}\left|R_2\right|\end{array}$