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InterviewSolution
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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) |
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Answer» Correct Answer - `8km//s` 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)` |
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