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A particle is oscillating on the $\mathrm{X}$-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:
(A) The image executes periodic motion
(B) The image executes non-periodic motion
(C) The turning points of the image are asymmetric w.r.t the image of the point at $x$ $=10 \mathrm{~cm}$
(D) The distance between the turning points of the oscillation of the image is $\frac{100}{21}$
Options:
Identify the correct statements:
(A) The image executes periodic motion
(B) The image executes non-periodic motion
(C) The turning points of the image are asymmetric w.r.t the image of the point at $x$ $=10 \mathrm{~cm}$
(D) The distance between the turning points of the oscillation of the image is $\frac{100}{21}$
Solution:
1328 Upvotes
Verified Answer
The correct answer is:
(A), (C), (D)
(A), (C), (D)
When object is at $8 \mathrm{~cm}$
Image $V_1=\frac{\mathrm{f} \times \mathrm{u}}{\mathrm{u}-\mathrm{f}}=\frac{5 \times 8}{8-5}=-\frac{40}{3} \mathrm{~cm}$
When object is at $12 \mathrm{~cm}$
Image $\mathrm{V}_2=\frac{\mathrm{f} \times \mathrm{u}}{\mathrm{u}-\mathrm{f}}=\frac{5 \times 12}{12-5}=-\frac{60}{7} \mathrm{~cm}$
Separation $=\left|\mathrm{V}_1-\mathrm{V}_2\right|=\frac{40}{3}-\frac{60}{7}=\frac{100}{21} \mathrm{~cm}$
So $\mathrm{A}, \mathrm{C}$ and $\mathrm{D}$ are correct statements.
Image $V_1=\frac{\mathrm{f} \times \mathrm{u}}{\mathrm{u}-\mathrm{f}}=\frac{5 \times 8}{8-5}=-\frac{40}{3} \mathrm{~cm}$
When object is at $12 \mathrm{~cm}$
Image $\mathrm{V}_2=\frac{\mathrm{f} \times \mathrm{u}}{\mathrm{u}-\mathrm{f}}=\frac{5 \times 12}{12-5}=-\frac{60}{7} \mathrm{~cm}$
Separation $=\left|\mathrm{V}_1-\mathrm{V}_2\right|=\frac{40}{3}-\frac{60}{7}=\frac{100}{21} \mathrm{~cm}$
So $\mathrm{A}, \mathrm{C}$ and $\mathrm{D}$ are correct statements.
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