Question

The position and velocity of a point object $A$ at a given instant of time are as shown. At that instant, considering the refraction through lens,

• A. The image is located on the principal axis at a distance of $40\text{cm}$ from lens.
• B. The speed of the image is $\sqrt{30}{\text{cms}}^{-1}$ w.r.t lens.
• C. The velocity components for image w.r.t lens are $2{\text{cms}}^{-1}$ along -ve X axis and $6{\text{cms}}^{-1}$ along -ve Y axis respectively.
• D. The velocity components for image w.r.t lens are $6{\text{cms}}^{-1}$ along -ve X axis and $2{\text{cms}}^{-1}$ along -ve Y axis respectively

Answer 1

Answer: (A), (C)

The correct options are
A The image is located on the principal axis at a distance of $40\text{cm}$ from lens.
C The velocity components for image w.r.t lens are $2{\text{cms}}^{-1}$ along -ve X axis and $6{\text{cms}}^{-1}$ along -ve Y axis respectively.

Consider the situation with sign convention $f=20\text{cm},u=-40\text{cm}\Rightarrow v=+40\text{cm}$

Velocity of object w.r.t ground
$={\overrightarrow{V}}_{OG}=\left(2{\text{cms}}^{-1}\right)\hat{i}+\left(2{\text{cms}}^{-1}\right)\hat{j}$
Velocity of lens w.r.t ground,
$={\overrightarrow{V}}_{LG}=\left(4{\text{cms}}^{-1}\right)\hat{i}-\left(4{\text{cms}}^{-1}\right)\hat{j}$
Let velocity of image w.r.t lens be
$={\overrightarrow{V}}_{IL}={\left(V_{IL}\right)}_x\hat{i}+{\left(V_{IL}\right)}_y\hat{j}$
Along X-axis
${\left(V_{IL}\right)}_x=\left(\frac{v^2}{u^2}\right){\left(V_{OL}\right)}_x=\left[\frac{{\left(40\right)}^2}{{\left(40\right)}^2}\right][2-4]=-2{\text{cms}}^{-1}\hat{i}$
Along Y-axis
${\left(V_{IL}\right)}_y=m{\left(V_{OL}\right)}_y=\left(\frac{v}{u}\right){\left(V_{OL}\right)}_y=\left(\frac{40}{-40}\right)[2-(-4\left)\right]=-\left[6{\text{cms}}^{-1}\right]\hat{\text{j}}$