A point object P  (at distance of 15 cm from lens) is approaching towards a thin and silvered equiconvex lens of local length 20 cm and refractive index 1.5 with a speed of  4 cms−1 as shown in figure. Find velocity of image.

1f=(μ−1)(1R+1R) ⇒120=(1.5−1)2R ∴R=20cmPower of equivalent mirror is P=2P1+Pm=220−1−10=110+110=15∴−1F=15 ∴F=−5 cm  Combination behaves like Concave Mirror  ∵1v+1u=1For1v−115=−15 Or  1v=115−15=1−315=−215 ∴v=−152cm ∴vimage=−v2u2dudt=−(−152)2(−15)2×4=−1cms−1