A) At angle \[45{}^\circ \] with the principal axis in downward direction.
B) At angle \[75{}^\circ \] with the principal axis in downward direction.
C) At angle \[45{}^\circ \] with the principal axis in upward direction.
D) At angle \[75{}^\circ \] with the principal axis in upward direction.
Correct Answer: A
Solution :
[a] The image of the object is formed on the other side of the lens at a distance of 20 cm. \[\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\] \[\Rightarrow \,\,\,\,\,\,-\frac{1}{{{v}^{2}}}\frac{dv}{dt}=-\frac{1}{{{u}^{2}}}\frac{du}{dt}\] \[\Rightarrow \,\,\,\,\frac{dv}{dt}={{\left( \frac{v}{u} \right)}^{2}}\frac{du}{dt}\] When \[v=u;\,\,\frac{dv}{dt}=\frac{du}{dt}\] \[\therefore \,\,\,\,\frac{{{y}_{1}}}{{{y}_{0}}}=\frac{v}{u}\] \[\frac{d{{y}_{1}}}{dt}=\frac{v}{u}\frac{d{{y}_{0}}}{dt}+{{y}_{0}}\frac{d}{dt}\left( \frac{v}{u} \right)\] In given position \[{{y}_{0}}=0\] \[\therefore \,\,\,{{V}_{YI}}=\left( \frac{v}{u} \right){{V}_{Y0}}=\left( \frac{20}{-20} \right){{V}_{Y0}}=-{{V}_{Y0}}\] \[\therefore \] Velocity of image will also make \[45{}^\circ \] with the principal axis (in downward direction).
You need to login to perform this action.
You will be redirected in
3 sec
