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# The velocity of the image when the object crosses the mean position and goes towards Q is A. $A\omega$B. $-A\omega$C. $\dfrac{A\omega }{2}$D. None of the above

Last updated date: 18th Sep 2024
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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.,
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

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
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
Now, using the given values in the above formula
\begin{align} & \Rightarrow \dfrac{1}{v}-\dfrac{1}{-3f}=\dfrac{1}{f} \\ \end{align}
\begin{align} & \Rightarrow \dfrac{1}{v}+\dfrac{1}{3f}=\dfrac{1}{f} \\ \end{align}
\begin{align} & \Rightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{3f} \\ \end{align}
$\Rightarrow \dfrac{1}{v}=\dfrac{2}{3f}$
So, we have
$\Rightarrow v=\dfrac{3f}{2}$
Now, using the formula for the magnification
$m=\dfrac{h'}{h}=\dfrac{v}{u}$
$\Rightarrow m=\dfrac{\dfrac{3f}{2}}{-3f}$
$\Rightarrow m=-\dfrac{1}{2}$
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,
$\Rightarrow y=\dfrac{A}{2}\sin (\omega t+\pi )$
On differentiating with respect to time
$\Rightarrow v=\dfrac{dy}{dt}=\dfrac{A\omega }{2}\cos (\omega t+\pi )$
When the image crosses the mean position, t = 0
So, velocity at that time will be
$\Rightarrow v=\dfrac{A\omega }{2}\cos \pi$
$\Rightarrow v=-\dfrac{A\omega }{2}$
Therefore, The velocity of the image when the object crosses the mean position and goes towards Q will be $-\dfrac{A\omega }{2}$

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,
$\Rightarrow v=\dfrac{A\omega }{2}\cos (\omega t+\pi )$
This formula will be useful in solving many such questions with ease.