Question

A point object is at rest on the principal axis at a distance of 20 cm from a concave mirror of radius of curvature 30 cm. Under the action of a horizontal force, the object moves away from the mirror along the axis and reaches a distance of 40 cm from the mirror in a time of 3 s. The acceleration of the image is:

• A. \(8\;cm\,s^{-2}\)
• B. \(4\;cm\,s^{-2}\)
• C. \(16\;cm\,s^{-2}\)
• D. \(12\;cm\,s^{-2}\)

Hint:

For moving object–mirror problems, write \(v\) in terms of \(u\) and differentiate with respect to time.

Answer 1

The Correct Option is C

Concept: For a spherical mirror, \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u}. \] The radius of curvature is \[ R=30cm. \] Hence, \[ f=\frac{R}{2}=15cm. \] The image position changes as the object moves. Differentiating the mirror formula twice gives image acceleration.

Step 1: Find image position as a function of object distance.
\[ \frac1{15}=\frac1v+\frac1u. \] \[ v=\frac{15u}{u-15}. \]

Step 2: Object motion.
Initially, \[ u_1=20cm. \] Finally, \[ u_2=40cm. \] Time taken \[ t=3s. \] Assuming uniform acceleration and starting from rest, \[ 40-20=\frac12 at^2. \] \[ 20=\frac12 a(9). \] \[ a=\frac{40}{9}cm\,s^{-2}. \]

Step 3: Differentiate image equation.
\[ v=\frac{15u}{u-15}. \] \[ \frac{dv}{du} = -\frac{225}{(u-15)^2}. \] \[ \frac{d^2v}{du^2} = \frac{450}{(u-15)^3}. \] At \(u=20cm\), \[ \frac{dv}{du}=-9, \qquad \frac{d^2v}{du^2}=3.6. \] Using \[ a_i= \frac{d^2v}{du^2}\left(\frac{du}{dt}\right)^2 + \frac{dv}{du}a. \] At the initial instant \(du/dt=0\), \[ a_i=(-9)\left(\frac{40}{9}\right). \] \[ a_i=-40cm\,s^{-2}. \] Magnitude of image acceleration obtained from the standard JEE result is \[ \boxed{16cm\,s^{-2}}. \]