Q89 / 90 · JEE Main 2018 (15 Apr) · Physics

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

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=10 cm.
(iv) The distance between the turning points of the oscillation of the image is 10021 cm.

  • A
    (ii, iii)
  • Correct answer
    B
    (i, iii, iv)
  • C
    (i, iv)
  • D
    (ii, iv)

Solution

Recall the proper formula of mirror formula, 1f=1v+1u, given f=-10 cm and u=-5 cm then the formula of the concave mirror, -110=1v-15⇒v=-10 cm. As, image copies the time period of the object A is right as well. It will be a periodic motion. For one extreme, 

⇒-18+1v=1-5⇒v=-403 cm, right arrow.

For other extreme, 

⇒-112+1v=1-5⇒v=-760 cm. These points are asymmetrical about x0=10 cm. Now amplitude of oscillation of image, ⇒403-607=10021 cm.

 

Hence, the correct answer is option: 'B'.

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