A point object is located at a distance of 100 m 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 50cm and spring constant 800 N/m as shown. Mass of the stand with lens is 2 kg. How much 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? (Neglect width of the stand).

8 kgm/s.

Let the distance of the lens from the object be $l$ when a real image is formed on the screen. Then $\frac{1}{100-l}=\frac{1}{-l}=\frac{1}{23}$ solving, we get $l = (50 ± 10\sqrt{2}) cm$.

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

∴ Phase difference between the two positions of real image must be π/2. As the two positions are symmetrically located about the origin, phase difference of any of these positions from origin must be π/4.

$⇒10\sqrt{2} cm = A \sin \frac{π}{4} ⇒ A = 20 cm$

To achieve this velocity at the mean position

$v_0 = Aω = A\sqrt{\frac{K}{m}}$

∴ Required impulse

$p = mv_0 = A\sqrt{Km }= 8 kgm/s.$