A point object is located at a distance of 100cm from a screen. A lens of focal length 23 cm mounted on a movable frictionless stad is kept between the source and the screen. The stand is attached to a spring of natureal length 50cm and spring constant `800N//m` as shown in Figure . 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? Also find this time gap. (Neglect width width of the stand)

Answer

`8km//s`

Explanation

Let the difference of th elens from the object be `l` when a real image is formed on the screen.
Then, `(1)/(100-l)-(1)/(-l)=(1)/(23)`
On solving, we get `l=(50+-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.
Therefore, phase difference between the two positions of real image must be `(pi)/(2)` . As the two positions are symmetrically located about the origin, phase difference of any of these positions from origin must be `(pi)/(4)`.
`rArr 10sqrt(2)cm=A "sin" (pi)/(5)rArr A=20cm`
To achieve this velocity at the mean position,
`v_(0)=A_(omega)=Asqrt((K)/(m))`
`:.` Required impulse `rho=mv_(0)=Asqrt(Km)=8kms^(-1)`