Question

A point source of light is moving at a rate of \[2cm/s \] towards a thin convex lens of focal length $10cm$ along its optical axis. When the source is \[15{{ }}cm\] away from the lens the image is moving at:
\[\left( {{A}} \right)4cm/s\] towards the lens
\[\left( {{B}} \right)8cm/s\] towards the lens
\[\left( {{C}} \right)4cm/s\] away from the lens
\[\left( {{D}} \right)8cm/s\] away from the lens

Answer 1

Hint: Use the given information about source distance and focal length in the relation between the distance of the object, the distance of image, and the focal length of the lens to calculate the distance of the image from the lens.
We may get the velocity of the object as well as the image by doing a derivative their distances from the lens w.r.t time.

Formula used:
\[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\]
Where \[u\] = the distance of the source from the lens.
\[v\] = the distance of the image from the lens.
\[f\]= the focal length of the lens
Since the position of the source is away from the lens,\[u\] is negative.
Hence the relation will be, \[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\] …………………(1)
By derivating the eq. (1) w.r.t time we get,
\[ - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} + \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}} = 0\] ………………….(2)

Complete step by step answer:
A point source is kept away from a convex lens of focal length \[f\] at a distance \[u\], and the image is situated at a distance \[v\]. The lense formula is,
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\] …………………(1)
Given, \[u\]=\[ - 15{{ }}cm\] [Since the position of the source is away from the lens,\[u\] is negative.]
\[f\]=$10cm$
\[\therefore \dfrac{1}{v} - \dfrac{1}{{( - 15)}} = \dfrac{1}{{10}}\]
\[ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{10}} - \dfrac{1}{{15}}\]
\[ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{30}}\]
\[ \Rightarrow v = 30\]
The velocity of the source is, \[\dfrac{{du}}{{dt}}\] and the velocity of the image is \[\dfrac{{dv}}{{dt}}\].
So, By derivating the eq. (1) w.r.t time we get,
\[ - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} + \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}} = 0\] ………………….(2)
Given, \[\dfrac{{du}}{{dt}} = 2cm - {s^{ - 1}}\]
Putting the values in eq.(2) we get,
\[\Rightarrow - \dfrac{1}{{{{30}^2}}}\dfrac{{dv}}{{dt}} + \dfrac{1}{{{{15}^2}}} \times 2 = 0\]
\[ \Rightarrow \dfrac{1}{{{{30}^2}}}\dfrac{{dv}}{{dt}} = \dfrac{1}{{{{15}^2}}} \times 2\]
\[ \Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{{30 \times 30 \times 2}}{{15 \times 15}}\]
\[ \Rightarrow \dfrac{{dv}}{{dt}} = 8\]
$\therefore $ The image is moving at a speed of \[8cm/s\] away from the lens.

Hence, the right answer is in option (D).

Note: The sign has been taken as if the direction shows left to right of a lens, the sign will be positive and if the direction shows right to left the sign will be negative. The right side of the lens is taken as the front side and the left of the lens is taken as the backside of the lens.
The position of the object is away from the lens that means the direction of light will be from the right to the left of the lens that’s why we take the negative sign.
The answer we get is that the velocity of the image is in a positive sign, which means the image is going from left to right of the lens i.e. away from the lens.