Answer:
Explanation:
1/v - 1/u = 1/f
given, f = 0.3 m, u = -0.4m
so, 1/v - 1/-0.4 = 1/0.3
or, 1/v = 1/0.3 - 1/0.4 = 1/1.2
v = 1.2 m
now, differentiating 1/v - 1/u = 1/f with respect to t,
-1/v² dv/dt + 1/u² du/dt = 0
or, dv/dt = (v/u)² du/dt
putting, du/dt = 0.01 m/s , v = 1.2 m and u = -0.4 m
so, dv/dt = (1.2/-0.4)² × 0.01
= 0.09 m/s
hence, speed of image with respect to lens is 0.09 m/s .
from formula of magnification
magnification, m = v/u
differentiating with respect to time both sides,
dm/dt = (u dv/dt - vdu/dt)/u²
= (-0.4 × 0.09 - 1.2 × 0.01)/(-0.4)²
= (-0.036 - 0.012)/0.16
= -0.048/0.16
= -0.3 m/s
hence, magnitude of rate of change of lateral magnification is 0.3 m/s
