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m = m o m e , size 12{m=m rSub { size 8{o} } m rSub { size 8{e} } } {}

where m o is the magnification of the objective and m e is the magnification of the eyepiece. This equation can be generalized for any combination of thin lenses and mirrors that obey the thin lens equations.

Overall magnification

The overall magnification of a multiple-element system is the product of the individual magnifications of its elements.

Microscope magnification

Calculate the magnification of an object placed 6.20 mm from a compound microscope that has a 6.00 mm focal length objective and a 50.0 mm focal length eyepiece. The objective and eyepiece are separated by 23.0 cm.

Strategy and Concept

This situation is similar to that shown in [link] . To find the overall magnification, we must find the magnification of the objective, then the magnification of the eyepiece. This involves using the thin lens equation.

Solution

The magnification of the objective lens is given as

m o = d i d o ,

where d o size 12{d rSub { size 8{o} } } {} and d i size 12{d rSub { size 8{i} } } {} are the object and image distances, respectively, for the objective lens as labeled in [link] . The object distance is given to be d o = 6.20 mm , but the image distance d i is not known. Isolating d i , we have

1 d i = 1 f o 1 d o , size 12{ { {1} over {d rSub { size 8{i} } } } = { {1} over {f rSub { size 8{o} } } } - { {1} over {d rSub { size 8{o} } } } } {}

where f o size 12{f rSub { size 8{o} } } {} is the focal length of the objective lens. Substituting known values gives

1 d i = 1 6 . 00 mm 1 6 . 20 mm = 0 . 00538 mm . size 12{ { {1} over {d rSub { size 8{i} } } } = { {1} over {6 "." "00 mm"} } - { {1} over {6 "." "20 mm"} } = { {0 "." "00538"} over {"mm"} } } {}

We invert this to find d i size 12{d rSub { size 8{i} } } {} :

d i = 186 mm. size 12{d rSub { size 8{i} } ="186 mm"} {}

Substituting this into the expression for m o size 12{m rSub { size 8{o} } } {} gives

m o = d i d o = 186 mm 6.20 mm = 30.0.

Now we must find the magnification of the eyepiece, which is given by

m e = d i d o , size 12{m rSub { size 8{e} } = - { {d rSub { size 8{i} } rSup { size 8{'} } } over {d rSub { size 8{o} } rSup { size 8{'} } } } } {}

where d i size 12{d rSub { size 8{i} rSup { size 8{'} } } } {} and d o size 12{d rSub { size 8{o} rSup { size 8{'} } } } {} are the image and object distances for the eyepiece (see [link] ). The object distance is the distance of the first image from the eyepiece. Since the first image is 186 mm to the right of the objective and the eyepiece is 230 mm to the right of the objective, the object distance is d o = 230 mm 186 mm = 44.0 mm . This places the first image closer to the eyepiece than its focal length, so that the eyepiece will form a case 2 image as shown in the figure. We still need to find the location of the final image d i in order to find the magnification. This is done as before to obtain a value for 1 / d i size 12{ {1} slash {d rSub { size 8{i} rSup { size 8{'} } } } } {} :

1 d i = 1 f e 1 d o = 1 50.0 mm 1 44.0 mm = 0.00273 mm . size 12{ { {1} over {d rSub { size 8{i} } rSup { size 8{'} } } } = { {1} over {f rSub { size 8{e} } } } - { {1} over {d rSub { size 8{o} } rSup { size 8{'} } } } = { {1} over {"50" "." "0 mm"} } - { {1} over {"44" "." "0 mm"} } = - { {0 "." "00273"} over {"mm"} } } {}

Inverting gives

d i = mm 0 . 00273 = 367 mm . size 12{d rSub { size 8{i} } rSup { size 8{'} } = - { {"mm"} over {0 "." "00273"} } = - "367 mm"} {}

The eyepiece’s magnification is thus

m e = d i d o = 367 mm 44 . 0 mm = 8 . 33 . size 12{m rSub { size 8{e} } = - { {d rSub { size 8{i} } rSup { size 8{'} } } over {d rSub { size 8{o} } rSup { size 8{'} } } } = - { { - "367 mm"} over {"44" "." "0 mm"} } =8 "." "33"} {}

So the overall magnification is

m = m o m e = ( 30.0 ) ( 8 . 33 ) = 250 . size 12{m=m rSub { size 8{o} } m rSub { size 8{e} } = \( - "30" "." 0 \) \( 8 "." "33" \) = - "250"} {}

Discussion

Both the objective and the eyepiece contribute to the overall magnification, which is large and negative, consistent with [link] , where the image is seen to be large and inverted. In this case, the image is virtual and inverted, which cannot happen for a single element (case 2 and case 3 images for single elements are virtual and upright). The final image is 367 mm (0.367 m) to the left of the eyepiece. Had the eyepiece been placed farther from the objective, it could have formed a case 1 image to the right. Such an image could be projected on a screen, but it would be behind the head of the person in the figure and not appropriate for direct viewing. The procedure used to solve this example is applicable in any multiple-element system. Each element is treated in turn, with each forming an image that becomes the object for the next element. The process is not more difficult than for single lenses or mirrors, only lengthier.

Questions & Answers

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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