# Concept of relative velocity and optics

**Question**

Find the velocity of image after !!1!! second in ground frame.

Radius of surface !!= 40 \ cm!!

a) !!- 4/3 \ m//s!!

b) !! 4/3 \ m//s!!

c) !!-17/3 \ m//s!!

d) !!17/3 \ m//s!!

**Solution**

We will use the gerneral equation for refraction at a spherical surface to find image distance,

!!mu_2/v- mu_1/u = (mu_2-mu_1)/R!!

!!=> 1.5/v-1/(-2)= (1.5-1)/(0.4)!!

!!=>v = 2 \ m!!

Differentiate the refraction formula with respect to time, !!t!!

!!-mu_2/(v^2)xx (dv)/(dt)+mu_1/(u^2)xx (du)/(dt)=0 .....(1)!!

!!(dv)/dt= V_(I//s)!! and !!(du)/(dt)= V_(O //s) = -2 \ m//s!!

By solving the equation !!(1)!!, we get

!!=> V_(I//s)= -4/3 \ m//s!!

Velocity of image with respect to ground,

!!V_(I//g)= 7+(-4/3)= 17/3 \ m//s!!