Question

The image of an object placed 3.2 cm from a convex
mirror has a transverse magnification of +0.4. Find:

(a) the location of the image
(b) the focal length.​

Answer 1

Answer:

A::C

Explanation:

f=0.3m u=−0.4m

Using lens formula

1v−1−0.4=10.3⇒v=1.2m

Now we have 1v−1u=1f, differentiationg w.r.t.t

we have −1v2dvdt+1u2dudt=0 given dudt=0.01m/simplies((dv)/(dt))=(1.20)^2/((0.4)^2)xx0.01=0.09m/sSo,rateofseparationoftheima≥(w.r.t.the≤ns)Now,m=(v)/(u)implies(dm)/(dt)=((udv)/(dt)-(vdu)/(dt))/(u^2)=(-(0.4)(0.09)-(1.2)(0.1))/((0.4)^2)=-0.35s^(-1)

Magnitude of rate of change of lateral magnification=0.35s−1