JEE Advanced 2022 Paper 2 · Q08 · Geometrical Optics

Question

An object and a concave mirror of focal length $f = 10\,\text{cm}$ both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_0 = 15\,\text{cm s}^{-1}$ 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 $u = 30\,\text{cm}$, 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 _____ $\text{cm s}^{-1}$.

[Figure: A concave mirror moves with velocity V_m to the right; an object at distance u to the left of the mirror moves to the right with velocity V_0. Both motions along the principal axis.]

SolveKar AI · Accepted Answer

Let positions of object and mirror in the lab frame be $x_o$ and $x_m$, with $u = x_m - x_o$. Mirror equation (object distance and image distance measured from the mirror): $1/v - 1/u = 1/f$ (using sign convention where distances are taken as magnitudes for a real image in front of the concave mirror): with $u = 30, f = 10$, $v = 15$ cm. Image position in lab frame: $x_i = x_m - v$. Time derivative: $\dot{x}_i = \dot{x}_m - \dot{v}$. From $1/v = 1/u - 1/f$, differentiating with respect to $u$: $-\dot{v}/v^2 = -\dot{u}/u^2$, so $\dot{v} = (v/u)^2 \dot{u} = (15/30)^2 \dot{u} = \dot{u}/4$. Here $\dot{u} = \dot{x}_m - \dot{x}_o$. Object moves toward the mirror in the lab (say $\dot{x}_o = +15$ cm/s rightward, mirror velocity $\dot{x}_m$ in same coordinate). $\dot{u} = \dot{x}_m - 15$. Image at rest: $\dot{x}_i = 0 \Rightarrow \dot{x}_m = \dot{v} = (\dot{x}_m - 15)/4$, giving $4\dot{x}_m = \dot{x}_m - 15$, $\dot{x}_m = -5$ cm/s. Magnitude $|V_m| = 5$ cm/s — but with the standard sign convention used in the official solution where $u$ is taken positive on the object side: $1/v + 1/u = 1/f$ gives $v = uf/(u-f) = 15$. Differentiating $1/v + 1/u = 1/f$: $\dot{v}/v^2 = -\ … [truncated]

JEE Advanced 2022 Paper 2 · Q08 · Geometrical Optics

Solution