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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

How is the de Broglie wavelength of electrons related to the quantization of their orbits in atoms and molecules?
Larissa Reply
How do you convert 0.0045kgcm³ to the si unit?
EDYKING Reply
how many state of matter do we really have like I mean... is there any newly discovered state of matter?
Falana Reply
I only know 5: •Solids •Liquids •Gases •Plasma •Bose-Einstein condensate
Thapelo
Alright Thank you
Falana
Which one is the Bose-Einstein
James
can you explain what plasma and the I her one you mentioned
Olatunde
u can say sun or stars are just the state of plasma
Mohit
but the are more than seven
Issa
what the meaning of continuum
Akhigbe Reply
What state of matter is fire
Thapelo Reply
fire is not in any state of matter...fire is rather a form of energy produced from an oxidising reaction.
Xenda
Isn`t fire the plasma state of matter?
Walter
all this while I taught it was plasma
Victor
How can you define time?
Thapelo Reply
Time can be defined as a continuous , dynamic , irreversible , unpredictable quantity .
Tanaya
unpredictable? but I can say after one o'clock its going to be two o'clock predictably!
Victor
what is the relativity of physics
Paul Reply
How do you convert 0.0045kgcm³ to the si unit?
flint
What is the formula for motion
Anthony Reply
V=u+at V²=u²-2as
flint
S=ut+½at
flint
they are eqns of linear motion
King
S=Vt
Thapelo
v=u+at s=ut+at^\2 v^=u^+2as where ^=2
King
hi
Mehadi
hello
King
Explain dopplers effect
Jennifer Reply
Not yet learnt
Bob
Explain motion with types
Bob
Acceleration is the change in velocity over time. Given this information, is acceleration a vector or a scalar quantity? Explain.
Alabi Reply
Scalar quantity Because acceleration has only magnitude
Bob
acleration is vectr quatity it is found in a spefied direction and it is product of displcemnt
bhat
its a scalar quantity
Paul
velocity is speed and direction. since velocity is a part of acceleration that makes acceleration a vector quantity. an example of this is centripetal acceleration. when you're moving in a circular patter at a constant speed, you are still accelerating because your direction is constantly changing.
Josh
acceleration is a vector quantity. As explained by Josh Thompson, even in circular motion, bodies undergoing circular motion only accelerate because on the constantly changing direction of their constant speed. also retardation and acceleration are differentiated by virtue of their direction in
fitzgerald
respect to prevailing force
fitzgerald
What is the difference between impulse and momentum?
Manyo
Momentum is the product of the mass of a body and the change in velocity of its motion. ie P=m(v-u)/t (SI unit is kgm/s). it is literally the impact of collision from a moving body. While Impulse is the product of momentum and time. I = Pt (SI unit is kgm) or it is literally the change in momentum
fitzgerald
Or I = m(v-u)
fitzgerald
Calculation of kinetic and potential energy
dion Reply
K.e=mv² P.e=mgh
Malia
K is actually 1/2 mv^2
Josh
what impulse is given to an a-particle of mass 6.7*10^-27 kg if it is ejected from a stationary nucleus at a speed of 3.2*10^-6ms²? what average force is needed if it is ejected in approximately 10^-8 s?
John
speed=velocity÷time velocity=speed×time=3.2×10^-6×10^-8=32×10^-14m/s impulse [I]=∆momentum[P]=mass×velocity=6.7×10^-27×32×10^-14=214.4×10^-41kg/ms force=impulse÷time=214.4×10^-41÷10^-8=214.4×10^-33N. dats how I solved it.if wrong pls correct me.
Melody
what is sound wave
Nworu Reply
sound wave is a mechanical longitudinal wave that transfers energy from one point to another
Ogor
its a longitudnal wave which is associted wth compresion nad rearfractions
bhat
what is power
PROMISE Reply
it's also a capability to do something or act in a particular way.
Kayode
Newton laws of motion
Mike
power also known as the rate of ability to do work
Slim
power means capabilty to do work p=w/t its unit is watt or j/s it also represents how much work is done fr evry second
bhat
what does fluorine do?
Cheyanne Reply
strengthen and whiten teeth.
Gia
a simple pendulum make 50 oscillation in 1minute, what is it period of oscillation?
Nansing Reply
length of pendulm?
bhat
Practice Key Terms 4

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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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