Question

The velocity of the image when the object crosses the mean position and goes towards Q is

A. $A\omega$
B. $-A\omega$
C. ${\displaystyle \frac{A\omega }{2}}$
D. None of the above

Answer 1

Hint: This is a case of a convex lens. In this situation, the image will be executing the SHM. The object held close to the convex lens of the focal length f, executes SHM between P and Q, O being the mean position. Take x-axis as the main axis of the lens and A < D to answer the following questions.

Formula used:
For solving this question, we will be using the formula for lenses, i.e.,
${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$

Complete step-by-step answer:
Before solving the given question, let us take a look at all the given parameters,
u = -3f
f = f
Since it’s a convex lens, the sign of the object distance is taken with negative sign
Now,
Applying the lens formula, we have
${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
Now, using the given values in the above formula
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{-3f}}={\displaystyle \frac{1}{f}}\end{array}$$
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{3f}}={\displaystyle \frac{1}{f}}\end{array}$$
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f}}-{\displaystyle \frac{1}{3f}}\end{array}$$
$$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{2}{3f}}$$
So, we have
$$\Rightarrow v={\displaystyle \frac{3f}{2}}$$
Now, using the formula for the magnification
$m={\displaystyle \frac{h^{\prime }}{h}}={\displaystyle \frac{v}{u}}$
$\Rightarrow m={\displaystyle \frac{{\displaystyle \frac{3f}{2}}}{-3f}}$
$\Rightarrow m=-{\displaystyle \frac{1}{2}}$
So, we can say that image will be inverted and have a path difference of $\pi$
SHM will be executed by the image in this case,
$\Rightarrow y={\displaystyle \frac{A}{2}}\sin (\omega t+\pi )$
On differentiating with respect to time
$$\Rightarrow v={\displaystyle \frac{dy}{dt}}={\displaystyle \frac{A\omega }{2}}\cos (\omega t+\pi )$$
When the image crosses the mean position, t = 0
So, velocity at that time will be
$\Rightarrow v={\displaystyle \frac{A\omega }{2}}\cos \pi$
$\Rightarrow v=-{\displaystyle \frac{A\omega }{2}}$
Therefore, The velocity of the image when the object crosses the mean position and goes towards Q will be $-{\displaystyle \frac{A\omega }{2}}$

So, the correct answer is “Option D”.

Note: You can see that the velocity of an object executing SHM can be given by the expression,
$$\Rightarrow v={\displaystyle \frac{A\omega }{2}}\cos (\omega t+\pi )$$
This formula will be useful in solving many such questions with ease.