A point is located at a distance 100 cm from the screen. A lens of focal length 23 cm mounted on a movable frictionless stand is kept between the object and screen. The stand is attached to massless spring of natural length 50 cm and spring constant `800(N)/(m)` as shown in the figure. Mass of the stand with lens is 2 kg. Initially spring is in its natural length. Now a sharp impulse J is given to the stand. Answer the following questions.
Q. Choose the correct options (s):

Options

A. Impulse J imparted to the stand is `8kg(m)/(sec)`
B. Impulse J imparted to the stand is `6kg(m)/(sec)`
C. difference between exreme position of lens is 40 cm
D. Difference between extreme position of lens is 60 cm

Answer

A, C

Explanation

`f=+23cm`
`(1)/(100-x)+(1)/(x)=(1)/(23)`
`implies(100)/(100x-x^(2))=(1)/(23)`
`x^(2)-100x+2300=0`
`x=(100+-sqrt(10000-9200))/(2)=50+-10sqrt(2)`
`x=(50-10sqrt(2))cm,(50+10sqrt(2))cm`.
Mean position is `x=50` real images will be formed when x
`=(50-10sqrt(2))cm` or `x=(50+10sqrt(2))cm`. since time difference between two consecutive real image formation on screen in same and this repeats four times in an oscillation, so displacement from mean position i.e., `Deltax=10sqrt(2)=(A)/(sqrt(2))`.
Here A is amplitude
`impliesA=20cm`
`t_(0)=((pi)/(2))/(omega)=(pi)/(2omega)`
`omega=sqrt((k)/(m))=sqrt((800)/(2))=20`
`t_(0)=(pi)/(40)sec`
Impulse imparted `=J=momegaA`
`=2.20(1)/(5)=8N-S`