JEE Advanced 2022 — Paper 2 — PHY-I solution

JEE Advanced2022Paper 2PHY-I

An object and a concave mirror of focal length $f = 10$ cm both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_{0} = 15$ cm s $^{- 1}$ towards the mirror with respect to a laboratory frame. The distance between the object and the mirror at a given moment is denoted by $u$ . When $u = 30$ cm, the speed of the mirror $V_{m}$ is such that the image is instantaneously at rest with respect to the laboratory frame, and the object forms a real image. The magnitude of $V_{m}$ is _____ cm s $^{- 1}$ .

Your answer

Solution

The linear magnification is $m = \frac{f}{u - f} = \frac{10}{30 - 10} = \frac{1}{2}$ (for a concave mirror with $u = 30$ cm).

For axial motion, velocity of the image with respect to the mirror equals $m^{2}$ times the velocity of the object with respect to the mirror (with the sign that makes a real image move opposite to the object). With the image at rest in the lab frame: $V_{image, lab} = V_{m} + m^{2} (V_{0} - V_{m}) = 0$ is not the standard sign convention; using the standard relation for axial mirror motion in the lab frame, $(V_{0} - V_{m}) m^{2} = V_{m}$ , which gives $V_{0} = V_{m} (1 + \frac{1}{m ^{2}}) = 5 V_{m}$ since $1/ m^{2} = 4$ .

Therefore $V_{m} = \frac{V _{0}}{5} = \frac{15}{5} = 3$ cm s $^{- 1}$ .

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