A point object is located at a distance of 100 cm from a screen. A lens of focal length 23 cm mounted on a movable frictionless stand is kept between the source and the screen. The stand is attached to a spring of natural length 50 cm and spring constant 800 N/m as shown. Mass of the stand with lens is 2 kg. How much minimum impulse P( in kg m/s) should be imparted to the stand so that a real image of the object is formed on the screen four times during a complete oscillation. All these are separated by same time interval. (Neglect width of the stand).

Let the distance of the lens from the object be $\mathcal{l}$

When a real image is formed on the screen.

 Then

$\frac{1}{100-\mathrm{\ell }}-\frac{1}{-\mathrm{\ell }}=\frac{1}{23}$

On solving, we get $\mathrm{\ell }=(50\pm 10\sqrt{2})$ cm

Now, if the lens performs SHM and a real image is formed after a fixed time gap, then this time gap must be one-fourth of the time period.

$\therefore$Phase difference between the two positions of real image must be $\frac{\mathrm{\pi }}{2}$. As the two positions are symmetrically located about the origin, phase difference of any of these positions from origin must be $\frac{\mathrm{\pi }}{4}$.

$\Rightarrow 10\sqrt{2}\mathrm{cm}=\mathrm{Asin}\frac{\mathrm{\pi }}{4}\Rightarrow \mathrm{A}=20\mathrm{cm}$

To achieve this velocity at the mean position,

${\mathrm{v}}_0=\mathrm{A\omega }=\mathrm{A}\sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$

$\therefore$Required impulse $\mathrm{p}={\mathrm{mv}}_0=\mathrm{A}\sqrt{\mathrm{K}/\mathrm{m}}=8\mathrm{kgm}/\mathrm{s}$