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JEE Advanced · Physics · 24. Ray Optics

Image of an object approaching a convex mirror of radius of curvature \(20 \mathrm{~m}\) along its optical axis is observed to move from \(\frac{25}{3} \mathrm{~m}\) to \(\frac{50}{7} \mathrm{~m}\) in \(30 \mathrm{~s}\). What is the speed of the object in \(\mathrm{km} \mathrm{h}^{-1}\) ?

  1. A 2
  2. B 3
  3. C 4
  4. D 5
Verified Solution

Answer & Solution

Correct Answer

(B) 3

Step-by-step Solution

Detailed explanation

Using mirror formula twice,
\(\frac{1}{+25 / 3}+\frac{1}{-u_1}=\frac{1}{+10}\)
or \(\quad \frac{1}{u_1}=\frac{3}{25}-\frac{1}{10}\) or \(u_1=50 \mathrm{~m}\)
and, \(\quad \frac{1}{(+50 / 7)}+\frac{1}{-u_2}=\frac{1}{+10}\)
\(\therefore \quad \frac{1}{u_2}=\frac{7}{50}-\frac{1}{10}\) or \(u_2=25 \mathrm{~m}\)
Speed of object \(=\frac{u_1-u_2}{\text { time }}\)
\(=\frac{25}{30} \mathrm{~ms}^{-1}\)
\(=3 \mathrm{kmh}^{-1}\)
\(\therefore\) The answer is 3 .
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