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A point source of light is moving at a rate of \[2\,cm - {s^{ - 1}}\] towards a thin convex lens of focal length 10 cm along its optical axis. When the source is 15 cm away from the lens, the image is moving at
A. \[4\,cm - {s^{ - 1}}\] towards the lens
B. \[8\,cm - {s^{ - 1}}\] towards the lens
C. \[4\,cm - {s^{ - 1}}\] away from the lens
D. \[8\,cm - {s^{ - 1}}\] away from the lens

seo-qna
Last updated date: 18th Jun 2024
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Answer
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403.5k+ views
Hint: Use lens formula to determine the image distance.
Differentiate the lens formula to determine the formula for rate of movement of image with respect to time.

Formula used:
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Here, f is the focal length, v is the image distance and u is the object distance.

Complete step by step answer:
We know that the focal length of the convex lens is positive. Also, the object distance from the lens is taken as negative.
We have the lens formula,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Here, f is the focal length, v is the image distance and u is the object distance.
Rearrange the above equation to determine the image distance as follows,
\[v = \dfrac{{fu}}{{u + f}}\]
Substitute 10 cm for f and \[ - 15\,cm\] for u in the above equation.
\[v = \dfrac{{\left( {10\,cm} \right)\left( { - 15\,cm} \right)}}{{ - 15\,cm + 10\,cm}}\]
Differentiate the lens formula with respect to time t as follows,
\[0 = - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} + \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}\]
\[ \Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} = \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}\]
\[\therefore \dfrac{{dv}}{{dt}} = \dfrac{{{v^2}}}{{{u^2}}}\dfrac{{du}}{{dt}}\]
Substitute \[ + 30\,cm\] for v, \[ - 15\,cm\] for u and \[ + 2\,cm - {s^{ - 2}}\] for \[\dfrac{{du}}{{dt}}\] in the above equation.
 \[\dfrac{{dv}}{{dt}} = \dfrac{{{{\left( {30\,cm} \right)}^2}}}{{{{\left( { - 15\,cm} \right)}^2}}}\left( {2\,cm - {s^{ - 2}}} \right)\]
 \[\dfrac{{dv}}{{dt}} = \left( 4 \right)\left( {2\,cm - {s^{ - 2}}} \right)\]
\[\dfrac{{dv}}{{dt}} = + 8\,cm - {s^{ - 2}}\]
The positive sign in the above equation implies that the image is moving along the positive direction of the x-axis which is away from the lens

So, the correct answer is “Option D”.

Note:
Read the properties of the image formed by the convex lens. The focal length of the convex lens is positive whereas the focal length of the concave lens is negative. The object distance or source distance is negative while the image distance is positive.