A) 50 cm
B) 100 cm
C) 200 cm
D) 400 cm
Correct Answer: B
Solution :
Key Idea: The radius of curvature of plane surface of plano-convex lens is \[\infty \] (infinite). Lens maker's formula for focal length of lens is, \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ...(i) We know that for plano-convex lens, the radius of curvature of plane surface is infinite, i.e.,. \[{{R}_{2}}=\infty \] Given, \[{{R}_{1}}=60\,\,cm,\,\,\mu =1.6\] Substituting the given values in Eq. (i), we get. \[\frac{1}{f}=(1.6-1)\left( \frac{1}{60}-\frac{1}{\infty } \right)=0.6\times \frac{1}{60}\] \[\therefore \] \[f=\frac{60}{0.6}=100\,cm\]
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