A) 1 m, 0.5 m, 0.25 m
B) 0.5 m, 1 m, 0.25 m
C) 0.5 m, 0.25 m, 1m
D) 0.25 m, 1m, 0.5 m
Correct Answer: D
Solution :
For surface P, \[\frac{1}{{{v}_{1}}}=\frac{1}{f}-\frac{1}{u}=1-\frac{1}{3}=\frac{2}{3}\]Þ \[{{v}_{1}}=\frac{3}{2}m\] For surface Q, \[\frac{1}{{{v}_{2}}}=\frac{1}{f}-\frac{1}{u}=1-\frac{1}{5}=\frac{4}{5}\]Þ \[{{v}_{2}}=\frac{5}{4}m\] \ \[{{v}_{1}}-{{v}_{2}}=0.25m\] Magnification of \[P=\frac{{{v}_{1}}}{u}=\frac{3/2}{3}=\frac{1}{2}\] \ Height of \[P=\frac{1}{2}\times 2=1m\] Magnification of \[Q=\frac{{{v}_{2}}}{u}=\frac{5/4}{5}=\frac{1}{4}\] \ Height of \[Q=\frac{1}{4}\times 2=0.5m\]
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