A) 10 cm
B) 20 cm
C) 40 cm
D) 80 cm
Correct Answer: B
Solution :
When a ray falls on convex surface of a plano-convex lens, then it is first refracted and reflected from plane surface and then finally refracted from convex surface. Thus, two refractions and one reflection take place. So, focal length of plano-convex lens is \[\frac{1}{F}=\frac{2}{{{f}_{l}}}+\frac{1}{{{f}_{m}}}\] ?(i) If plane surface is silvered, so \[{{f}_{m}}=\frac{{{R}_{2}}}{2}=\frac{\infty }{2}=\infty \] Now, \[\frac{1}{{{f}_{1}}}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[=(\mu -1)\left( \frac{1}{R}-\frac{1}{\infty } \right)\] \[=\frac{(\mu -1)}{R}\] \[\therefore \] \[\frac{1}{F}=\frac{2(\mu -1)}{R}+\frac{1}{\infty }\] \[=\frac{2(\mu -1)}{R}\] or \[F=\frac{R}{2(\mu -1)}\] ?(ii) Given, \[R=20\,cm,\,\mu =1.5\] Hence, \[F=\frac{20}{2(1.5-1)}\] \[=\frac{20}{2\times 0.5}=20\,cm\]
You need to login to perform this action.
You will be redirected in
3 sec
