A particle is oscillating on the \(X\)-axis with an amplitude \(2\ \mathrm{cm}\) about the point \(x_0=10\ \mathrm{cm}\), with a frequency \(\omega\). A concave mirror of focal length \(5\ \mathrm{cm}\) is placed at the origin (see figure). Identify the correct statements.
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Find the image positions for the extreme object positions \(8\ \mathrm{cm}\) and \(12\ \mathrm{cm}\) and for \(10\ \mathrm{cm}\) using \(\frac1v+\frac1u=\frac1f\) with \(f=-5\ \mathrm{cm}\).
The turning points of the image are asymmetric w.r.t. the image of the point at \(x=10\ \mathrm{cm}\).
The distance between the turning points of the oscillation of the image is \(\frac{100}{21}\ \mathrm{cm}\).
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The Correct Option isA
Solution and Explanation
The object oscillates between \(u=-8\ \mathrm{cm}\) and \(u=-12\ \mathrm{cm}\) (distances from the mirror) about \(-10\ \mathrm{cm}\); \(f=-5\ \mathrm{cm}\). Mirror formula \(\frac1v=\frac1f-\frac1u\): \(u=-8\): \(\frac1v=-\frac15+\frac18=-\frac{3}{40}\), \(v=-\frac{40}{3}=-13.33\ \mathrm{cm}\). \(u=-12\): \(\frac1v=-\frac15+\frac1{12}=-\frac{7}{60}\), \(v=-\frac{60}{7}=-8.57\ \mathrm{cm}\). \(u=-10\): \(\frac1v=-\frac15+\frac1{10}=-\frac1{10}\), \(v=-10\ \mathrm{cm}\). (1) The object returns to every position after one period, so the image does too: the image motion is periodic (but not simple harmonic). Statement (1) is true and (2) is false. (3) Turning points at \(13.33\) and \(8.57\ \mathrm{cm}\) are \(3.33\ \mathrm{cm}\) and \(1.43\ \mathrm{cm}\) from the image point \(10\ \mathrm{cm}\): asymmetric. True. (4) Distance between turning points \(=\frac{40}{3}-\frac{60}{7}=\frac{280-180}{21}=\frac{100}{21}\ \mathrm{cm}\). True. So statements (1), (3) and (4) are all correct; statement (2) is the only false one. Choosing the first correct statement as the single answer. Answer: (1) The image executes periodic motion.
