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A lens placed at the origin with its axis pointing along the x axis produces a real inverted image at $x = - 24 cm$ that is twice as tall as the object.

What is the image distance?

Why is the image distance $s' = 24\text{ cm}$ not $-24\text{ cm}$ (actually already given in the question)?

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Use the direction of the light rays from image towards lens or mirror to be +ve direction in any problem to face no further problems. –  ABC Mar 30 '13 at 5:53
    
.yes the distance can be used as $-24cm$ but just keep the +ve direction same for measuring every distance. –  ABC Mar 30 '13 at 6:00

1 Answer 1

up vote 2 down vote accepted

I would say that the distance is a scalar quantity, and therefore an absolute value. So, the image is AT -24cm, but is located AT A DISTANCE of 24cm. So the position is negative, but distance is always positive.

If you want to know the new Cartesian Sign Convention:

http://www.it.iitb.ac.in/ekshiksha/images_light_10/figure_9.JPG

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icic, so in the magnification or lens equation formula, I should be using displacement (vector) right? $\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}$ and $m = \frac{s'}{s}$. The the negative value? –  Jiew Meng Mar 30 '13 at 7:15
    
yes, that is what the sign conventions are for –  Saurabh Raje Mar 30 '13 at 7:57

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