A) \[100\,\,cm\]
B) \[200\,\,cm\]
C) \[178\,\,cm\]
D) \[150\,\,cm\]
Correct Answer: A
Solution :
If \[{{R}_{1}}\] and \[{{R}_{2}}\] are the radii of curvature of first and second refracting surfaces of a thin lens with optical centre \[C\] of focal length \[f\] and refractive index \[\mu \] then according to lens Maker's formula \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] where \[\mu \] is refractive index of material of lens with respect to surrounding medium. For plano-convex lens, \[{{R}_{1}}=50\,\,cm,\,\,{{R}_{2}}=\infty \] (for plain surface) \[\frac{1}{f}=(1.5-1)\left( \frac{1}{50}-\frac{1}{\infty } \right)\] or \[\frac{1}{f}=0.5\times \frac{1}{50}\] or \[f=\frac{50}{0.5}=100\,\,cm\]
You need to login to perform this action.
You will be redirected in
3 sec
