question_answer

A rod of length 10cm lies along the principal axis of a concave mirror of focal length 10cm in such a way that its end closer to the pole is 20cm away from the mirror length of image is:

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 \[=\left| {{v}_{A}} \right|-\left| {{V}_{B}} \right|\] \[=(20-10)\,cm\,=5cm\]