A particle is oscillating on the $x$-axis with an amplitude $2\ \text{cm}$ about the point $x_0 = 10\ \text{cm}$ with a certain frequency. A concave mirror of focal length $5\ \text{cm}$ is placed at the origin, facing the particle.
Identify the correct statements.
(i) The image executes periodic motion.
(ii) The image executes non-periodic motion.
(iii) The turning points of the image are asymmetric with respect to the image of the point at $x = 10\ \text{cm}$.
(iv) The distance between the turning points of the oscillation of the image is $\dfrac{100}{21}\ \text{cm}$.
Answer: (B) (i), (iii), (iv)
For a real object at distance $u$ in front of a concave mirror of focal length $f = 5\ \text{cm}$, the image distance is
$$v = \frac{uf}{u - f} = \frac{5u}{u - 5}$$
The object moves between $u = 8\ \text{cm}$ and $u = 12\ \text{cm}$:
$$u = 8:\ v = \frac{40}{3} = 13.33\ \text{cm}, \qquad u = 10:\ v = 10\ \text{cm}, \qquad u = 12:\ v = \frac{60}{7} = 8.57\ \text{cm}$$
- The image position is a single-valued function of the object position, so it repeats with the same period: the motion is periodic. (i) is true and (ii) is false.
- The turning points are $3.33\ \text{cm}$ and $1.43\ \text{cm}$ from the image of $x = 10\ \text{cm}$, so they are asymmetric. (iii) is true.
- The distance between them is $\dfrac{40}{3} - \dfrac{60}{7} = \dfrac{280 - 180}{21} = \dfrac{100}{21}\ \text{cm}$. (iv) is true.
The correct statements are (i), (iii), (iv).
Solution by Sreeraj P, M.Sc Physics