question_answer

The radius of the convex surface of plan convex lens is 20 cm and the refractive index of the material of the lens is 1.5. The focal length is:                                                                                          [BHU M-2003]

A)  \[30\,\,\,cm\]                                  

B)  \[50\,\,\,cm\]

C)  \[20\,\,\,cm\]                                  

D)  \[25\,\,\,cm\]

Correct Answer: D

Solution :

                 Key Idea: Radius of plane surface of lens is infinity. From the lens formula focal length of a lens is \[\frac{1}{f}=\left( \mu -1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]          Where \[\mu \] is refractive index of the material of the lens and \[{{R}_{1}}\] and \[{{R}_{2}}\] are radii of curvatures of lens surface. Given, \[\mu =1.5,\,{{R}_{1}}=20\,cm,\,{{R}_{2}}=\infty \] \[\left( \text{plane}-\text{surface} \right)\]                 \[\therefore \]                  \[\frac{1}{f}=\left( 1.5-1 \right)\left( \frac{1}{20}-\frac{1}{\infty } \right)\]                 \[\Rightarrow \]                               \[\frac{1}{f}=\frac{1}{40}\]                 \[\Rightarrow \]                              \[f=40\,cm\].