WEBVTT
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An object is placed 30 centimetres from a diverging lens.
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If the distance between the lens and the image is six centimetres, what is the magnification? a) one-third, b) one-fifth, c) five, d) three, e) one-fifteenth.
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The problem gives us the object distance of 30 centimetres.
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๐ subscript o is used to represent object distance.
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We are also told that the image distance is six centimetres.
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๐ subscript i is used to represent image distance.
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The problem asks us to solve the magnification of our lens.
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The letter ๐ is used to represent magnification.
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To help us visualise our problem better, letโs draw in our object, lens, and image.
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Because the problem did not state what the object look like, we chose an arrow to represent the general case.
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The yellow arrow represents the object and is at 30 centimetres from the lens.
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The pink arrow represents the image and is at six centimetres from the lens.
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The green lens depicts a diverging lens because itโs thinner in the middle and thicker on the ends.
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The problem did not state which side of the lens the object was placed.
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We chose the left side of the lens to represent the general case.
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Before solving the problem mathematically, we can eliminate some answer choices by recalling a few facts about diverging lenses.
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Any lens that causes parallel light rays to diverge after refracting is called the diverging lens.
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A diverging lens always produces a virtual image.
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This is why we drew our image on the same side of the lens as our object to represent that it is virtual.
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We can now go back and add in a negative sign in front of our image distance to make it negative six centimetres to show that this is a virtual image.
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A diverging lens always produces an upright image relative to the object.
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In our diagram, we chose to draw our object arrow pointing up.
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Therefore, image arrow is also pointing up.
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On the other hand, if our object arrow is pointing down, then our image arrow would also be pointing down.
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A diverging lens always produces a diminished image.
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This is why our diagram shows an image arrow that is smaller than the object arrow.
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From a magnification standpoint, if a lens produces an image that is diminished, then it has a magnification of less than one.
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Looking at our answer choices since both c and d are greater than one, we can cancel them out.
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Recall that the magnification is equal to the ratio of the image height to the object height as well as the negative ratio of the image distance to the object distance.
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Because the problem did not give us the height of either the image or the object, but did give us the distance of both the image and the object, weโll use the second equation to solve for the magnification.
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Substituting in the values from our problem, we get that the magnification is equal to the negative ratio of negative six centimetres divided by 30 centimetres.
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Because magnification is a ratio, we can say that the centimetres in the numerator cancel out with the centimetres in the denominator, leaving no units.
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Multiplying out the numerator, we get negative one times negative six is six.
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And in the denominator, we get 30.
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So our magnification is six divided by 30.
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Because six goes into 30 five times, reducing the fraction of six divided by 30, we get one divided by five or one-fifth.
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The final answer is that the magnification is one-fifth or answer choice b.