Question

A projectile is fired at t = 0 horizontally from height H infront of a convex lens as shown in figure, then at t = 0

• A. Speed of image of projectile is same as that of projectile
• B. Speed of image of projectile is more than that of projectile
• C. Speed of image is not horizontal
• D. Image of projectile is moving away from lens.

Answer 1

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

The correct options are
B Speed of image of projectile is more than that of projectile
C Speed of image is not horizontal
D Image of projectile is moving away from lens.

For the lens,
$u=-2f$ and by using the formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, $v$ turns ut to be $2f$.
And by $f^{-1}=v^{-1}-u^{-1}\Rightarrow 0=\frac{-1}{v^2}\frac{dv}{dt}+\frac{1}{u^2}\frac{du}{dt}$

$V_{Ix}=\frac{v^2}{u^2}{V}_{Ox}$

$V_{Ix}={\left(\frac{-2f}{2f}\right)}^2{V}_{Ox}=V_{Ox}\dots \left(1\right)$

By $\frac{I}{O}=\frac{v}{u}\Rightarrow I=O\left(\frac{v}{u}\right)$ and $O=H,\frac{dO}{dt}=0$

$V_{Iy}=\frac{dI}{dt}=O\frac{d}{dt}\left(\frac{v}{u}\right)+\frac{v}{u}\frac{dO}{dt}=H\frac{d}{dt}\left(\frac{v}{u}\right)\ne 0\dots \dots \left(2\right)$

From (1) and (2), the speed of the image is more than the speed of the object. And the speed of the image is making some angle with the horizontal. The horizontal component of the image's velocity is positive, so it is moving away from the lens.