Question:
When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of 9 cms\(^{-1}\), the speed (in cms\(^{-1}\)) with which image moves at that instant is ____.
When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of 9 cms\(^{-1}\), the speed (in cms\(^{-1}\)) with which image moves at that instant is ____.
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Differentiate the mirror formula \(\frac1v+\frac1u=\frac1f\) with respect to time: \(\frac{dv}{dt} = -\frac{v^2}{u^2}\frac{du}{dt}\).
Updated On: Nov 5, 2026
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Correct Answer: 1
Solution and Explanation
Mirror formula: \(\dfrac1v+\dfrac1u = \dfrac1f\) with \(u=-30\) cm and \(v=-10\) cm (real image, in front of the mirror).
Differentiate with time: \(-\dfrac{1}{v^2}\dfrac{dv}{dt}-\dfrac{1}{u^2}\dfrac{du}{dt} = 0\), so \(\dfrac{dv}{dt} = -\dfrac{v^2}{u^2}\dfrac{du}{dt}\).
Speed of image \(= \dfrac{v^2}{u^2}\times\) speed of object \(= \dfrac{100}{900}\times 9 = 1\) cm/s.
Answer: 1
Differentiate with time: \(-\dfrac{1}{v^2}\dfrac{dv}{dt}-\dfrac{1}{u^2}\dfrac{du}{dt} = 0\), so \(\dfrac{dv}{dt} = -\dfrac{v^2}{u^2}\dfrac{du}{dt}\).
Speed of image \(= \dfrac{v^2}{u^2}\times\) speed of object \(= \dfrac{100}{900}\times 9 = 1\) cm/s.
Answer: 1
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