Question

A point object is projected from the principle axis in front of convex lens of focal length $12\text{cm}$ as shown in figure. Which of the following statements is/are correct?

• A. Speed of image perpendicular to principal axis is $4{\text{ms}}^{-1}$
• B. The angle that velocity of the image makes with the principal axis is greater than ${45}^{\circ }$
• C. Image of object moves away from lens
• D. Speed of image along principle axis is $7{\text{ms}}^{-1}$

Answer 1

Answer: (A), (B), (C)

Image of object moves away from lens

Given, $u=-30\text{cm};f=12\text{cm}{V}_{Ox}=8{\text{ms}}^{-1};V_{Oy}=6{\text{ms}}^{-1}$

Using lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}......\left(1\right)$

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{12}-\frac{1}{30}$

$\Rightarrow v=20\text{cm}$

Differentiating $\left(1\right)$ w.r.t $t$ on both sides,

$0=-\frac{1}{v^2}\frac{dv}{dt}+\frac{1}{u^2}\frac{du}{dt}$

$\frac{1}{v^2}\frac{dv}{dt}=\frac{1}{u^2}\frac{du}{dt}$

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

$\Rightarrow V_{Ix}={\left(\frac{20}{30}\right)}^2\times 8=\frac{32}{9}{\text{ms}}^{-1}$

From, magnification relation,

$m=\frac{I}{O}=\frac{v}{u}$

$\Rightarrow I=\frac{v}{u}O$

$\Rightarrow \frac{dI}{dt}=\left(\frac{v}{u}\right)\frac{dO}{dt}+O\frac{d\left(\frac{v}{u}\right)}{dt}[\because O=0\text{cm}]$

$\Rightarrow \frac{dI}{dt}=\left(\frac{v}{u}\right)\frac{dO}{dt}$

$\Rightarrow V_{Iy}=\left(\frac{v}{u}\right)V_{Oy}$

$\Rightarrow V_{Iy}=\left(\frac{20}{30}\right)\times 6=4{\text{ms}}^{-1}$

$\because \tan \theta =\frac{V_{Iy}}{V_{Ix}}=\frac{4\times 9}{32}=\frac{36}{32}>1$

$\Rightarrow \theta >{45}^{\circ }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right),\left(B\right)\&\left(C\right)$ is the correct answer.