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 Makers 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, \[{{R}_{1}}=50\,cm,\,R=\infty \] (for plane 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 \[f=\frac{50}{0.5}=100\,cm\]
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