<< Chapter < Page Chapter >> Page >
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

Calculate the Newton's the weight of a 2.5 Kilogram of melon. What is its weight in pound?
Rialyn Reply
calculate the tension of the cable when a buoy with 0.5m and mass of 20kg
Iga Reply
what is displacement
Nyamza Reply
it's the time rate of change of distance
Mollamin
distance in a given direction is diplacement
Musa
Distance in a spacified direction
Gift
you shouldn't say distance,displacement and distance are two different things .distance can be lopped curved but displacement is always in a straight line so you can't use distance to define it. displacement is the change of position in a specified direction.
Joshua
Well stayed josh👍
Gift
thank you gift.
Joshua
well explained
Mary
what is the meaning of physics
Alausa Reply
to study objects in motion and how they interact or take part in the natural phenomenon of the universe.
Phill
an object that has a small mass and an object has a large mase have the same momentum which has high kinetic energy
Faith Reply
The with smaller mass
Gift
how
Faith
Since you said they have the same momentum.. So meaning that there is more like an inverse proportionality in the quantities used to find the momentum. We are told that the the is a larger mass and a smaller mass., so we can conclude that the smaller mass had higher velocity as compared to other one
Gift
Mathamaticaly correct
megavado
Mathmaticaly correct :)
megavado
I have proven it by using my own values
Gift
Larger mass=4g Smaller mass=2g Momentum of both=8 Meaning V for L =2 and V for S=4 Now find there kinetic energies using the data presented
Gift
grateful soul...thanks alot
Faith
Welcome
Gift
2 stones are thrown vertically upward from the ground, one with 3 times the initial speed of the other. If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? If the slower stone reaches a maximum height of H, how high will the faster stone go
Julliene Reply
30s
Gift
how can i calculate it's height
Julliene
is speed the same as velocity
Faith Reply
no
Nebil
in a question i ought to find the momentum but was given just mass and speed
Faith
just multiply mass and speed then you have the magnitude of momentem
Nebil
Yes
Gift
Consider speed to be velocity
Gift
it worked our . . thanks
Faith
Distinguish between semi conductor and extrinsic conductors
Okame Reply
Suppose that a grandfather clock is running slowly; that is, the time it takes to complete each cycle is longer than it should be. Should you (@) shorten or (b) lengthen the pendulam to make the clock keep attain the preferred time?
Aj Reply
I think you shorten am not sure
Uche
shorten it, since that is practice able using the simple pendulum as experiment
Silvia
it'll always give the results needed no need to adjust the length, it is always measured by the starting time and ending time by the clock
Paul
it's not in relation to other clocks
Paul
wat is d formular for newton's third principle
Silvia
okay
Silvia
shorten the pendulum string because the difference in length affects the time of oscillation.if short , the time taken will be adjusted.but if long ,the time taken will be twice the previous cycle.
FADILAT
discuss under damped
Prince Reply
resistance of thermometer in relation to temperature
Ifeanyi Reply
how
Bernard
that resistance is not measured yet, it may be probably in the next generation of scientists
Paul
Is fundamental quantities under physical quantities?
Igwe Reply
please I didn't not understand the concept of the physical therapy
John Reply
physiotherapy - it's a practice of exercising for healthy living.
Paul
what chapter is this?
Anderson
this is not in this book, it's from other experiences.
Paul
am new in the group
Daniel
please I have probably with calculate please can you please and help me out
John Reply
Sure
Gift
What is Boyce law
Sly Reply
Boyles law states that the volume of a fixed amount of gas is inversely proportional to pressure acting on that given gas if the temperature remains constant which is: V<k/p or V=k(1/p)
FADILAT
Practice Key Terms 4

Get the best College physics course in your pocket!





Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask