A) 1 : 2
B) 3 : 1
C) 1 : 3
D) 2 : 1
Correct Answer: B
Solution :
Focal length of mirror \[f=\frac{R}{2}=\frac{10}{2}=5cm\] For part PQ : transverse magnification length of image L1 = \[\left( \frac{f}{f-u} \right)\times {{L}_{0}}\] = \[\left( \frac{-5}{-5-(-20)} \right)\times {{L}_{0}}=\frac{-{{L}_{0}}}{3}\] For part QR : longitudinal magnification Length of image \[{{L}_{2}}={{\left( \frac{f}{f-u} \right)}^{2}}{{L}_{0}}\] = \[{{\left( \frac{-5}{-5-(-20)} \right)}^{2}}\times {{L}_{0}}=\frac{{{L}_{0}}}{9}\] Þ \[\frac{{{L}_{1}}}{{{L}_{2}}}=\frac{3}{1}\]You need to login to perform this action.
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