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An object is 2.3 cm tall, and 5.5 cm from a converging lens.  The image is upright and...

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lkehoe | Valedictorian

Posted April 27, 2013 at 8:43 PM via web

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An object is 2.3 cm tall, and 5.5 cm from a converging lens.  The image is upright and 9.0 cm tall.  What is the focal length of the lens? (Using the magnification fo find the image distance)

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ishpiro | Teacher | (Level 1) Associate Educator

Posted April 28, 2013 at 1:53 AM (Answer #1)

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The magnification M is the ratio of image height to the object height:

`M=h_i/h_o=9/2.3=3.91`

This ratio is equal to the ratio of the image distance to the object distance, from the similarity of the appropriate triangles:

`d_i/d_o = 3.91`

From here we can find image distance:

`d_i=3.91*d_o=3.91*5.5=21.52 cm`

Now we can use the lens equation to find the focal distance. Since the image is upright, it means that this is a virtual image and it is located on the same side of the lens as the object. Thus, the image distance has to come into the equation with the negative sign:

`1/f=1/d_o - 1/d_i`

`1/f = 1/5.5 - 1/21.52=0.14`

f = 7.39 cm

The focal length of the lens is 7.39 cm.

Note that the object distance is less than the focal length, so the object is placed in between the lens and its focal point. This results in the image being virtual, as discussed above.

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