A) \[+\left( \frac{R}{U-2R} \right)v_{0}^{2}\]
B) \[-{{\left( \frac{R}{R-2U} \right)}^{2}}v_{0}^{{}}\]
C) \[-{{\left( \frac{R}{2U-2R} \right)}^{2}}v_{0}^{{}}\]
D) \[+\left( \frac{R}{2U-2} \right)v_{0}^{2}\]
Correct Answer: B
Solution :
[b] For concave mirror \[\frac{2}{R}=\frac{1}{v}+\frac{1}{u}\text{ or }\frac{2}{-R}=\frac{1}{v}+\frac{1}{-u}\] \[\therefore \frac{1}{v}=\frac{1}{U}-\frac{2}{R}=\frac{R-2U}{UR}\text{ or }v=\left[ \frac{RU}{R-2U} \right]\] In spherical mirror, image velocity \[{{v}_{t}}=-\left[ \frac{{{v}^{2}}}{{{u}^{2}}} \right]{{v}_{0}}=-{{\left[ \frac{RU}{R-2U} \right]}^{2}}\frac{{{v}_{0}}}{{{U}^{2}}}=-{{\left[ \frac{R}{R-2U} \right]}^{2}}{{v}_{0}}\]
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