A) 10cm
B) 15cm
C) 2.5cm
D) 5cm
Correct Answer: B
Solution :
[b] \[f=-10cm\] For end \[A,\text{ }{{\text{u}}_{A}}=-\,20\text{ }cm\] \[\frac{1}{{{v}_{A}}}+\frac{1}{{{u}_{A}}}=\frac{1}{f}\] \[\frac{1}{{{v}_{A}}}+\frac{1}{(-\,20)}=\frac{1}{-\,10}\] \[For\text{ }end\,B,\text{ }{{\text{u}}_{B}}=-\,30\text{ }cm\] Image position of end B \[\frac{1}{{{v}_{B}}}+\frac{1}{{{u}_{B}}}=\frac{1}{f}\] \[\frac{1}{{{v}_{B}}}+\frac{1}{(-\,30)}=\frac{1}{(-\,10)}\] \[{{V}_{B}}=-15cm\] Length of image \[=\left| {{v}_{A}} \right|-\left| {{V}_{B}} \right|\] \[=(20-10)\,cm\,=5cm\]You need to login to perform this action.
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