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Q. A point object is kept just beyond the first focus of a convex lens. If the lens starts moving towards the right with a constant velocity, the image will
Question

NTA AbhyasNTA Abhyas 2022

Solution:

$\frac{\text{dv}}{\text{dt}} = \left(\frac{\text{v}}{\text{u}}\right)^{2} \times \frac{\text{du}}{\text{dt}}$
$\left(\text{V}\right)_{\text{image/lens}} = \left(\frac{\text{v}}{\text{u}}\right)^{2} \times \left(\text{V}\right)_{\text{object/lens}}$
till $u$ become $2f$ , $m>-1$ so $V_{image/lens}>V_{object/lens}$ and the image will move towards left and when
$u$ become $>2f$ , $V_{image / lens} < V_{object / lens}$ and the image will move towards the right.