A point object (S) is located at a distance of 100cm 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. The maximum value of impulse P should be imparted to the stand so that a real image of the object is formed on the screen after a fixed time gap is x kg m/s. Then the value of x is (Neglect width of the stand)
 
 

Let the distance of the lens from the object be l when a real image is formed on the screen. Then  $\frac{1}{100-l}=\frac{1}{-l}=\frac{1}{23}$
On solving, we get  $l=(50\pm 10\sqrt{2})cm$
Now, if the lens performs SHM and real images is formed after a fixed time gap then this gap must be one fourth of the time period.
   Phase difference between the two positions of real image must be /2. As the two positions are symmetrically located about the origin, phase difference of any of these positions from origin must be /4.
 $\Rightarrow 10\sqrt{2}cm=A\sin (\pi /4)\Rightarrow A=20cm$
To achieve this velocity at the mean position  $v_0=A\omega =A\sqrt{\frac{K}{m}}$
$\therefore$  Required impulse