A) 10 cm
B) 20 cm
C) 30 cm
D) 45 cm
Correct Answer: A
Solution :
[a]| When the plane surface is silvered the focal length \[{{f}_{1}}\] is given by \[\frac{1}{{{f}_{1}}}=\frac{2(\mu -1)}{R}\] | ...(i) |
| But when the convex surface is silvered, the focal length \[{{f}_{2}}\] is given by \[\frac{1}{{{f}_{2}}}=\frac{2\mu }{R}\] | ...(ii) |
| Dividing Eq. (i) by Eq. (ii), we have\[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{\mu }{\mu -1}=\frac{1.5}{1.5-1}=3\] | |
| or \[{{f}_{2}}=\frac{{{f}_{1}}}{3}=\frac{30}{3}=10\,cm\] | |
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