Name: Rayleigh Scattering
Uploaded: 2019-07-19
Duration: 6 min 37 s
Description: Why is the sky so blue? Well, it is blue because of something called Rayleigh scattering. This phenomenon is named after Lord Rayleigh. Watch the video to le...

Question

A short object of length

L

is placed along the principal axis of a concave mirror away from focus. The object distance is

u

. If the mirror has a focal length

f

, what will be the length of the image? You may take

L < < |v - f|

.

Accepted Answer

As the mean distance of object from mirror is $u$

$f$ and $μ_{2} = (u + \frac{L}{2})$

Let the image of the two ends of object form at distance $v_{1}$ and $v_{2} (v_{1} > v_{2})$ . So the length of the image on the principal axis is $L' = (v_{1} - v_{2})$

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{v} = \frac{u - f}{u f} \Rightarrow v = \frac{u f}{u - f}$

So $L' = v_{1} - v_{2} = \frac{(u - \frac{L}{2}) f}{(u - \frac{L}{2}) - f} - \frac{(u + \frac{L}{2}) f}{(u + \frac{L}{2}) - f}$

$\Rightarrow L' = f \frac{u - \frac{L}{2}}{(u - f - \frac{L}{2})} - \frac{u + \frac{L}{2}}{(u - f + \frac{L}{2})}$

$\Rightarrow L' = f [\frac{(u - \frac{L}{2}) (u - f + \frac{L}{2}) - (u + \frac{L}{2}) (u - f - \frac{L}{2})}{(u - f - \frac{L}{2}) (u - f + \frac{1}{2})}]$

$= \frac{f [u^{2} - u f + \frac{u L}{2} - \frac{u L}{2} + \frac{f L}{2} - \frac{L^{2}}{4} - (u^{2} - u f - \frac{u L}{2} + \frac{u L}{2} - \frac{f L}{2} - \frac{L^{2}}{4})]}{{(u - f)}^{2} - \frac{L^{2}}{4}}$

$∵ L < < (u - f)$

$∴ \frac{L^{2}}{4} < < < (u - f)$

So neglecting the terms $\frac{L^{2}}{4}$

$L^{2} = f [\frac{\frac{f L}{2} + \frac{f L}{2}}{{(u - f)}^{2}}]$

$L' = \frac{f f L}{{(u - f)}^{2}} \Rightarrow L' = \frac{L f^{2}}{{(u - f)}^{2}}$

It is the length of image.

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