Ray Optics

JEE Main level, single correct, medium

47 of 295

Question 47 of 295

A particle is oscillating on the xx-axis with an amplitude 2cm2cm about the point x0=10cm{x}_{0}=10cm with a frequency. A concave mirror of focal length 5cm5cm is placed at the origin (see figure).

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=10cmX=10cm.
(iv) The distance between the turning points of the oscillation of the image is
10021cm\dfrac{100}{21}cm.

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Answer: (B) (i, iii, iv)

Recall the proper formula of mirror formula, 1f=1v+1u\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}, given f=−10cmf=-10cm and u=−5cmu=-5cm then the formula of the concave mirror, −110=1v−15⇒v=−10cm-\dfrac{1}{10}=\dfrac{1}{v}-\dfrac{1}{5}\Rightarrow v=-10cm. As, image copies the time period of the object (A)\left(A\right) is right as well. It will be a periodic motion. For one extreme,
⇒−18+1v=1−5⇒v=−403cm\Rightarrow -\dfrac{1}{8}+\dfrac{1}{v}=\dfrac{1}{-5}\Rightarrow v=-\dfrac{40}{3}cm, right arrow.
For other extreme,

⇒−112+1v=1−5⇒v=−760cm\Rightarrow -\dfrac{1}{12}+\dfrac{1}{v}=\dfrac{1}{-5}\Rightarrow v=-\dfrac{7}{60}cm. These points are asymmetrical about x0=10cm{x}_{0}=10cm. Now amplitude of oscillation of image, ⇒403−607=10021cm\Rightarrow \dfrac{40}{3}-\dfrac{60}{7}=\dfrac{100}{21}cm.

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