Question

A point object is projected from the principle axis in front of convex lens of focal length 12 cm as shown in figure then

• A. Speed of image along principle axis is 7 m/s
• B. Speed of image perpendicular to principle axis is 4 m/s
• C. Angle that velocity of image makes with principle axis is greater than ${45}^{\circ }$
• D. Image of object moves away from lens

Answer 1

Answer: (B), (C), (D)

The correct options are
B Speed of image perpendicular to principle axis is 4 m/s
C Angle that velocity of image makes with principle axis is greater than ${45}^{\circ }$
D Image of object moves away from lens

Using $f^{-1}=v^{-1}-u^{-1}$ $u=-30\Rightarrow v=20$
$f^{-1}=v^{-1}-u^{-1}$
By differentiating on both sides,
$0=-\frac{1}{v^2}\frac{dv}{dt}+\frac{1}{u^2}\frac{du}{dt}$

$V_{Ix}=\frac{v^2}{u^2}{V}_{Ox}\Rightarrow V_{Ix}={\left(\frac{20}{30}\right)}^2\times 8=\frac{32}{9}m/s$

By $m=\frac{I}{O}=\frac{v}{u}\Rightarrow I=\frac{v}{u}O\Rightarrow \frac{dI}{dt}=\left(\frac{v}{u}\right)\frac{dO}{dt}+O\frac{d\frac{v}{u}}{dt}$
$O=0\Rightarrow \frac{dI}{dt}=\left(\frac{v}{u}\right)\frac{dO}{dt}$
$\Rightarrow V_{Iy}=\left(\frac{v}{u}\right)V_{oy}\Rightarrow V_{Iy}=\left(\frac{20}{30}\right)\times 6=4m/s$

$\tan \theta =\frac{V_{Iy}}{V_{Ix}}=\frac{4\times 9}{32}=\frac{36}{32}>1\Rightarrow \theta >{45}^{\circ }$