question_answer

A object is moving with velocity \[\nu .\] (w. r. t. earth) parallel to plane mirror \[{{M}_{1}}.\]Another plane mirror \[{{M}_{2}}\] makes an angle \[\beta \] with the vertical as shown in figure. Then velocity of image in mirror \[{{M}_{2}}\] w.r.t. the image in \[{{M}_{1}}\] is -

A)  \[2\,\nu \sin \,\,\beta \]               

B) \[\nu \cos 2\,\,\beta \]

C) \[2\,\nu \sin \,\,\beta \]                

D) \[\nu \,\sqrt{2}\]

Correct Answer: C

Solution :

From image diagram

&

\[\Rightarrow \]   \[{{\nu }_{{{I}_{2}},\,\,{{I}_{1}}}}=\sqrt{{{(\nu -\nu \cos 2\beta )}^{2}}+{{(v\sin 2\beta )}^{2}}}\]

\[=2\,\,\nu \sin \beta \]