A particle executes a simple harmonic motion of amplitude 1.0 cm along the principle axis of a convex lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the lens amplitude of oscillation of image of the particle. Does the image also execute simple harmonic motion ?

Options

A. 1.5 cm
B. 2.3 cm
C. 2.8 cm
D. 3.2 cm

Answer

B

Explanation

The particle has an amplitude A= 1 cm
Its minimum dis"tan"ce from the lens =20-1=19 cm and the maximum dis"tan"ce =20+1=21 cm
`therefore " For " u_(1) c, (1)/(f)=(1)/(v)-(1)/(u)`
For a convex lens, `u(-ve), v(+ve) and f(+ve)`
`therefore (1)/(v_(1))=(1)/(f)+(1)/(u)=(1)/(12)-(1)/(19)=(7)/(19xx12)`
`therefore v_(1)=(19xx12)/(7)=32.58 cm`
Similarly for `u_(2)=21 cm`,
`(1)/(v_(2))=(1)/(12)-(1)/(21)=(9)/(21xx12)=(1)/(28)`
`therefore v_(2)=28 cm`
`therefore` Path length of the oscillation of the image
=32.58-28=4.58 cm
`therefore` Amplitude `=(4.58)/(2)=2.29=2.3 cm`