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, \[{{R}_{1}}=60\text{ }cm,~\,\,\mu =1.6\] |
| 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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