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JEE Advanced 2022 · Paper 2, Physics question 8

An object and a concave mirror of focal length f=10f = 10 cm both move along the principal axis of the mirror with constant speeds. The object moves with speed V0=15 cm s−1V_0 = 15\ \mathrm{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 uu. When u=30u = 30 cm, the speed of the mirror VmV_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 VmV_m is _____ cm s−1\mathrm{cm\,s^{-1}}.

An object (upright arrow) on the principal axis moving with velocity V0 towards a concave mirror on the right; the mirror moves with velocity Vm; u is the object-mirror distance.
Answer and worked solution

Official answer: 3

Image position. Use the sign convention with u=−30u=-30 cm and f=−10f=-10 cm:
1v=1f−1u=−110+130  ⟹  v=−15 cm.\frac1v=\frac1f-\frac1u=-\frac1{10}+\frac1{30}\implies v=-15\text{ cm}.
The image is real, with m=−vu=−12m=-\frac vu=-\frac12.

Velocities along the axis. In the mirror's frame, the image velocity is −m2-m^2 times the object velocity. Take the direction of the object's motion as positive, and let the mirror move with VmV_m. Relative to the mirror, the object moves at 15−Vm15-V_m, so the image moves at −14(15−Vm)-\frac14(15-V_m).

Image at rest in the lab.
Vm−14(15−Vm)=0  ⟹  Vm=3 cm s−1.V_m-\frac14(15-V_m)=0\implies V_m=3\ \mathrm{cm\,s^{-1}}.
So the mirror moves at 3 cm/s, in the same direction as the object.

Topics:Spherical mirrorsRelative velocity
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