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QJEEAdv22Phdea6: JEE Advanced 2022 Physics - Class 12 JEEAdv22Phdea6 Solution: An object and a concave mirror of focal length both move along…

An object and a concave mirror of focal length f=10 cmf=10 \mathrm{~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} \mathrm{~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=30 cmu=30 \mathrm{~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} \,\mathrm{s}^{-1} .

JEE Advanced 2022 Paper 2 Online Physics - Geometrical Optics Question 15 English

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JEE Advanced 2022 Paper 2 Online Physics - Geometrical Optics Question 15 English Explanation
Magnification,

amp;m=fu−famp;=1030−10=12\begin{aligned}& m =\frac{f}{u-f} \\\\& =\frac{10}{30-10}=\frac{1}{2}\end{aligned}

Also m2=vmv0−vm\quad m^2=\frac{v_m}{v_0-v_m} [where, vm=v_m= speed of mirror]

amp;m2(vo−vm)=vmamp;⇒(v0−vm)×14=vmamp;⇒v0=5vmamp;⇒vm=v05=155amp;⇒vm=3 cm/s\begin{aligned}& m^2\left(v_{\mathrm{o}}-v_{\mathrm{m}}\right) =v_m \\\\& \Rightarrow \left(v_0-v_m\right) \times \frac{1}{4} =v_{\mathrm{m}} \\\\& \Rightarrow v_0 =5 v_m \\\\& \Rightarrow v_m =\frac{v_0}{5}=\frac{15}{5} \\\\& \Rightarrow v_m =3 \mathrm{~cm} / \mathrm{s}\end{aligned}

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