• Tags: image-velocity, differentiation, mirrors
• Difficulty: Advanced

For a mirror, differentiating 1/v + 1/u = 1/f with respect to time:

-1/$v^{2}$ (dv/dt) - 1/$u^{2}$ (du/dt) = 0

Therefore: dv/dt = -(v/u)^{2} × (du/dt) = -$m^{2}$ × v_object

This means: v_image = $m^{2}$ × v_object (magnitude), and the image moves in the opposite direction to the object along the principal axis.

For an object moving perpendicular to the principal axis: v_image(perpendicular) = m × v_object(perpendicular).

JEE frequently asks problems where objects move along or perpendicular to the principal axis.