Question

Focal length of the convex lens shown in the figure is $20\text{cm}$. Find the position of the image and magnitude of image velocity, when the object is at a distance of $30\text{cm}$ from the pole.

• A. $30\text{cm},\sqrt{292}\text{mm/s}$
• B. $60\text{cm},\sqrt{190}\text{mm/s}$
• C. $60\text{cm},\sqrt{292}\text{mm/s}$
• D. $30\text{cm},\sqrt{192}\text{mm/s}$

Answer 1

The correct option is C $60\text{cm},\sqrt{292}\text{mm/s}$

From the data given in the diagram,
$u=-30\text{cm},f=+20\text{cm}$

Using the lens formula, we have :

$\frac{1}{v}-\frac{1}{-30}=\frac{1}{+20}$

Solving this equation we get, $v=+60\text{cm}$

Lateral magnification, $m=\frac{v}{u}=\frac{+60}{-30}=-2$

Component of velocity of object along the principal axis ,
$u_x=5\cos {37}^{\circ }=4\text{mm/s}$ (towards the lens)

Component of velocity of image along the principal axis,
$v_x=m^2\times 4\text{mm/s}=4\times 4=16\text{mm/s}$ (away from the lens).

Component of velocity of object perpendicular to the principal axis ,
$u_y=5\sin {37}^{\circ }=3\text{mm/s}$ (upwards).

$\therefore$ Component of velocity of image perpendicular to principal axis,
$v_y=m\left(3\text{mm/s}\right)=(-2)\left(3\text{mm/s}\right)=-6\text{mm/s}$
Or, $v_y=6\text{mm/s}$ (downwards)

These all points are shown in the figure given below.




$\therefore$ Image velocity $v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{(16{)}^2+(6{)}^2}=\sqrt{292}\text{mm/s}$

Hence, option $\left(c\right)$ is the correct alternative.