When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of 9 cms –1 , the speed (in cms –1 ) with which image moves at that instant is ____.

Solution

Physics sir · 3 steps

Differentiate the mirror formula: along the axis, image speed is object speed times (v/u)².

1. Differentiate in time. Start from 1/v + 1/u = 1/f. The focal length is fixed, so differentiating gives -1/v²dv/dt - 1/u²du/dt = 0.

   dv/dt = -v²/u² du/dt

2. Put in numbers. With sign convention, u = -30 cm and v = -10 cm, so v²/u² = 100/900 = 1/9. The object speed is 9 cm s⁻¹, so the image speed is 1/9 × 9 = 1 cm s⁻¹.
3. So. The image moves at 1 cm s⁻¹. The minus sign only tells you it moves in the opposite sense along the axis.

The common slip: 3. Using |v/u| = 1/3 without squaring it gives 3, not 1. Axial speed scales with m², not m.

The idea to keep. For motion along the principal axis, image speed = m² × object speed, where m = -v/u. The square matters, and the image always moves opposite to the object.

Answer from the NTA final answer key. Solution by PrepWiser.