Question Details

[answered] LensesGoals and IntroductionWhen a ray of light strikes the


LensesGoals and IntroductionWhen a ray of light strikes the surface of a material, some of the light may be reflected and sometransmitted into the material. The portion of the light ray that is transmitted into the materialundergoes refraction, based on the difference between the index of refraction of the material andthe incident medium. This refraction, or bending of the light ray will occur again when that lightray exits the material and transmits back into the surrounding medium. The refraction of the lightray at each boundary is governed by Snell?s Law, as you may have investigated in the Reflectionand Refraction of Light lab activity.If the material is in the shape of a rectangular block, then a light ray that strikes the side is likelyto exit the opposite side, traveling parallel to the original incident ray of light, but offset by acertain amount that would depend on the index of refraction of the material and the size of block.This is because the entrance and exit boundary surfaces are parallel to each other. This is itself aninteresting, measurable phenomenon, but of further interest is what happens when we change theshape of the entrance and exit surfaces of the material.If we curve the entrance and/or exit surfaces of the material so that they have a spherical shape,we will find that applying Snell?s Law at each boundary results in light undergoing a netrefraction either towards, or away from a line that passes through the center of the material. Wecall this line the principal axis. If the shape of the lens results in the light being bent towards theprincipal axis, we call this type a lens a converging lens. If the shape of the lens results in thelight being bent away from the principal axis, we call this type a lens a diverging lens.Figure 1 illustrates what happens to light rays that are parallel (from a very distant source) whenthey are focused, or refracted by a converging lens. This kind of diagram is called a ray diagram.We see that the rays cross the principal axis at a single point, called the focal point of the lens.The distance from the center of the lens to this point is called the focal length of the lens. Thefocal point is the place where an image, created by the lens, would form if the object, or source,is very far away from the lens (thus, making the incident light rays nearly parallel).Figure 2 illustrates what happens to light rays that are parallel (from a very distant source) whenthey are refracted by a diverging lens. In this ray diagram, we see that the light rays emerge as ifthey all had passed through a focal point that is behind the lens on the incident side. The distancefrom the center of the lens to this point is called the focal length of the lens. The focal point is theplace where an image would appear to have been created if we look through the lens towards theobject. The light our eye would collect would look like it was coming from the focal point!Figure 1Figure 2The behavior of incident light that is parallel to the principal axis through the center of the lens isestablished in the above ray diagrams, but what if the rays are coming from an object, or source,much nearer the lens? If we imagine light rays leaving in different directions from a single pointon the object, it is clear they will not hit the lens traveling parallel to the principal axis, as wasthe case in Figure 1 and 2. What we would like is to identify a set of predictive behaviors forsome of these rays, so that we can still locate where the emerging rays cross, or where theyappear to cross, when viewed by an observer to the right of the lens (we always draw our raydiagrams with the light traveling from left to right).To aid in developing rules for some rays and how they will behave when passing through thelens, we define a focal point on each side of the lens, where each is a focal length from the centerof the lens. For the purposes of trying to create a self-consistent set of rules, we place a ?prime?symbol on the right focal point for the converging lens and a ?prime? symbol on the left focalpoint for the diverging lens. The following can then be stated for the drawing of three rays,originating from the same point on the object, which will aid in locating the image (where theemerging rays cross, or appear to cross):The ray leaving a point on the object parallel to the principal axis is deflected so that itpasses through the focal point, F? of the lens (or travels in a direction as if it traveledthrough the focal point F?).The ray that goes through the focal point, F, or travels in an incident direction as if itwere going through the focal point F, emerges from the lens parallel to the principal axis.The ray directed at the center of the lens is undeflected.We can see an example of these rules applied to light leaving the top of an object with its bottomin line with the principal axis (Figure 3). This allows us to locate the image because light leavingthe bottom of the image would emerge along the principal axis, meaning the bottom of the imagewill be located on the principal axis also.Figure 3We can also draw these rays, following the same rules for the diverging lens (Figure 4). Notethat the ?primed? focal point is on the left for the case of the diverging lens, which means weinterpret the rules in a somewhat strange way. The incident ray that travels parallel to theprincipal axis should emerge through F?, or along a direction as if it went through F?. We had toback-trace the ray from the vertical line of action through the lens (where we can imagine thatnet refraction occurs ? it of course really occurs at the entrance and exit surfaces) until it hit F?and then extend the line forward. A similar thought process must be applied to the incident raythat travels ?as if it were going to go through the focal point F? in order to choose the properinitial direction before it emerges parallel to the principal axis. Also, note that because the lightFigure 4travels away from the principal axis after passing through a diverging lens, there is no possibleway for it to cross on the right side of the lens. In order to find the image, we must back-traceeach of the emerging rays to see where they appear to cross behind the lens.When the emerging rays actually cross on the right side of the lens, we say this is a real imagebecause the light is actually focused there and is present at that location. This means the imagecould be projected onto a screen. When the light appears to cross on the left side of the lens (suchas in the case of a diverging lens), we say this is a virtual image, and can only be seen by lookingback through the lens and seeing where the light appears to be coming from. If you were to placea screen there, you would not be able to get an image to appear on the screen because the light isnot actually passing through that location ? it just appears to be coming from there. There is aninteresting scenario where a converging lens can be used to produce a virtual image, instead of areal image. Do some research to understand how this could be possible!Now, in either case, we define the distance of the object from the center of the lens as so and thedistance of the image from the center of the lens as si (Figure 5). Recall that the focal length ofthe lens is the distance between a focal point and the center of a lens, f. We can also define theheight of the object as ho and the height of the image as hi. Realize that, for each of thesequantities, there is a sign convention, as defined in your textbook. For example, the focal lengthof a converging lens is quoted as a positive value, whereas the focal length of a diverging lens isquoted as a negative value. Please consult your textbook to become familiar with the signconventions for the other quantities, as this will be necessary to answer pre-lab questions inaddition to performing calculations in this lab activity.Figure 5There are two primary relationships between these quantities that apply to either lens when lgithpasses through them ? Eq. 1 and Eq. 2. Remember that there are sign conventions for each ofthese quantities that must be followed. The first equation is called the thin lens equation, since itapplies well when the lens material is thin. The second equation defines the magnification of theimage, m, which can be expressed in terms of a ratio of the image and object height, or in termsof the negative ratio of the image distance and object distance.1 1 1o i s s f? ?(Eq. 1)i io oh smh s? ? ?(Eq. 2)Lastly, it is possible to model the behavior of light as it passes through one lens and then asecond lens. Realize that the light that will enter the second lens in the series is what hasemerged from the first lens. This means that the second lens doesn?t ?see? the original object, but?sees? the image from the first lens. This means that the object distance for the second lens is thedistance between the image location from the first lens and the center of the second lens! Also,the height of the image from the first lens will be the object height for the second lens! The thinlens equation and the magnification can then be applied for the second lens to find the finalimage location and height.In today?s lab, you will investigate the focal length of a converging lens and confirm the thin lensequation by projecting an image on the screen. You will also investigate the criteria required forthe converging lens to create a virtual image, instead of a real image. Finally, you will attempt toconfirm the focal length of a diverging lens by using a multi-lens system.Goals: (1) Explore and investigate the properties of converging and diverging lenses(2) Measure and confirm the thin lens model(3) Utilize and solve a multi-lens system for the purposes of determining the focallength of a diverging lensProcedureEquipment ? mini-optics bench, light source, crossed-arrow object mask, three componentcarriers, 75 mm convex lens, 150 mm concave lens, screen, a windowNOTE: The room lights should be turned down during the rest of the experiment.1) Lay out the lab optics bench from the kit box. Note that the rulers on the sides are in units ofmm. You can place components of this lab on this bench during the experiment. Some aremagnetic and will be held in place when you place them on the bench.2) Place the 75 mm convex lens on a component carrier and then place the carrier on the bench.Place the screen on a carrier and then place that carrier on the bench, as well.NOTE: The word ?convex? here refers to the shape of the entrance and exit surfaces of the lens.This results in a converging lens, so remember that the 75 mm convex lens is a converging lens!3) Record the expected focal length of this converging lens, 75 mm.4) Go to a window, or in the hall so that you can have light from a very distant object passthrough the lens, and then strike the screen. Move the screen, or the lens, until you are able tocreate a focused image of the object on the screen. Measure and Record the distance betweenthe lens and the screen, which is the experimental focal length of the lens.Question 1: Why should the distance you just measured be representative of the focal length ofthe lens? Consider Figure 1 in your explanation. Is this image inverted or is does it have the sameorientation as the object? Explain this observation.5) At your table, place the light source on the optics bench at one end. Place the crossed-arrowobject mask onto a component carrier. Then place that carrier in front of the light source, upagainst the source. Place the lens carrier after that and finally the screen carrier.6) Measure and record the height of the vertical arrow on the crossed-arrow object mask. Thiswill be the initial object height during the experiment.7) Note the position of the crossed-arrow object mask along the optics bench. Move the lens sothat the distance between the object and the lens is about 10.0 cm. Measure and record theobject distance you have and then move the screen to locate what you feel is the location wherethe image is in focus. Measure and record the image distance, the distance between the lens andthe screen. Measure and record the height of the image on the screen. You may want to applyany sign conventions for these quantities, if necessary.8) Repeat step 7 using approximate object distances of 15.0 cm, 20.0 cm, and 25.0 cm. Then,construct a single ray diagram, for the converging lens in one of these scenarios.9) Now, move the lens so that it is 6.0 cm from the crossed-arrow object. Note that you are notable to find a real image on the screen. Look through the lens from the side where the screen waslocated. You will likely see the object still, because your eyes collect from behind the lens, butyou should also see a virtual image of the object within the lens. Record your observations aboutthis image ? is it magnified? Right-side up or upside down? Is it closer to the lens than the objector further away?Question 2: Why was a virtual image formed with the object at that distance from the lens?10) Place the 150 mm concave lens on a component carrier and place it between the crossedarrowmask carrier and the 75 mm lens carrier on the optics bench. Record the expected focallength of the diverging lens, -150 mm.NOTE: The word ?concave? here refers to the shape of the entrance and exit surfaces of the lens.This results in a diverging lens, so remember that the 150 mm concave lens is a diverging lens!11) Place the 150 mm lens about 4.0 cm after the crossed-arrow object, and then place the 75mm lens about 6.0 cm after the 150 mm lens. Finally, place the screen after the 75 mm lens andlocate a focused image by moving the screen back and forth. Measure and record the distancebetween the crossed-arrow object and the 150 mm lens (the first lens), the distance between thelenses, and the distance between the 75 mm lens (the second lens) and the location of the image.Note that the first value is the object distance for the first lens, and the last value is the imagedistance for the second lens. Also, measure and record the height of the image on the screen.This value is the image height for the second lens. You may want to apply any sign conventionsfor these quantities, if necessary.12) Try moving the second lens and the screen to find another set of distances where a focusedimage forms on the screen. Repeat the measurements of step 11 once you have created this setup.13) Try to construct a single ray diagram, for the multi-lens system with the diverging lens firstand the converging lens second.Question 3: Why can?t we repeat the procedure we used at the start of the lab to find the focallength of the diverging lens? Explain your answer.Question 4: Could we repeat the procedure in step 11 or 12 with the lenses reversed? Why orwhy not? In other words, is there a scenario where we could get an image to appear on the screenif the lenses are reversed? Feel free to try this, but either way, explain why, or why not, this ispossible.As always, be sure to organize your data records for presentation in your lab report, using tablesand labels where appropriate.Data AnalysisUse the data from steps 7 and 8 to calculate the focal length of the converging lens. Then,calculate the mean value of your results.Use the data from steps 7 and 8 to calculate the magnification and image height in each casewithout using the image height you measured experimentally.For the multi-lens system in step 11, use the thin lens equation for each lens, the data yougathered, and the expected focal length of the converging lens to calculate the focal length of thediverging lens. Be careful with all of your sing conventions!For the multi-lens system in step 12, use the thin lens equation for each lens, the data yougathered, and the expected focal length of the converging lens to calculate the focal length of thediverging lens. Be careful with all of your sing conventions!Calculate the mean value for the focal length of the diverging lens.Error AnalysisCalculate the percent error between the focal length you measured in step 4 and the expectedfocal length of the converging lens.experimental expectedexpected% 100%f ferrorfCalculate the percent error between the mean value of the focal length of the converging lens youfound in the Data Analysis section and the expected focal length of the converging lens.For each case in steps 7 and 8, calculate the percent error between each of the image height thatyou measured and the image height you calculated in the Data Analysis section.measured data analysisdata analysis% 100%h herrorhCalculate the percent error between the mean value of the focal length of the diverging lens youfound in the Data Analysis section and the expected focal length of the diverging lens.experimental expectedexpected% 100%f ferrorfQuestion 5: Is the thin lens equation accurate, as stated? What sources of error might beresponsible for deviations from this model? Explain your answer by referencing your results.Questions and ConclusionsBe sure to address Questions 1 through 5 and describe what has been verified and tested by thisexperiment. What are the likely sources of error? Where might the physics principlesinvestigated in this lab manifest in everyday life, or in a job setting?Pre-Lab QuestionsPlease read through all the instructions for this experiment to acquaint yourself with theexperimental setup and procedures, and develop any questions you may want to discuss withyour lab partner or TA before you begin. Then answer the following questions and type youranswers into the Canvas quiz tool for ?Lenses,? and submit it before the start of your lab sectionon the day this experiment is to be run.PL-1) When an image appears on a screen, we can assuredly say that the image isA) a virtual image.B) a real image.C) upside down.D) right-side up.PL-2) An object is placed 10.0 cm in front of a converging lens with a focal length of 5.0 cm.What is the image distance? Express your answer in cm.PL-3) An object is placed 10.0 cm in front of a diverging lens with a focal length of -5.0 cm.What is the image distance? Express your answer in cm.PL-4) When an object is placed in front of a diverging lens, the image is formed on the left sideof the lens (the same side as the object), and the image distance is negative. Suppose you havelight from an object pass through a diverging lens and then a converging lens that is 10.0 cmaway from the diverging lens. When writing the thin lens equation for the second lens (theconverging lens), the object distance isA) less than 10.0 cm.B) equal to 10.0 cmC) greater than 10.0 cm.D) 0 because the light doesn?t get there.PL-5) A diverging lens with focal length -15.0 cm has an object placed 5.0 cm in front of it. Aconverging lens with a focal length of 8.0 cm is placed on the other side of the diverging lens,10.0 cm away. How far from the converging lens will the final image be located? In other words,find the image distance for the converging lens. Express your answer in cm


Converging lens=75mm=7.5cm

 

Object

 

distance

 

(u) cm Object

 

distance u < 2F 12

 

10

 

8

 

U=2F

 

15

 

u > 2F

 

25

 

30

 

35

 

Mean focal length 1 1 ?1 ?1

 

/c m/c m

 

u v Image

 

distance(v)(cm) 1

 

Focal

 

/c m...

 


Solution details:
STATUS
Answered
QUALITY
Approved
ANSWER RATING

This question was answered on: Sep 18, 2020

PRICE: $15

Solution~0001013140.zip (25.37 KB)

Buy this answer for only: $15

This attachment is locked

We have a ready expert answer for this paper which you can use for in-depth understanding, research editing or paraphrasing. You can buy it or order for a fresh, original and plagiarism-free copy from our tutoring website www.aceyourhomework.com (Deadline assured. Flexible pricing. TurnItIn Report provided)

Pay using PayPal (No PayPal account Required) or your credit card . All your purchases are securely protected by .
SiteLock

About this Question

STATUS

Answered

QUALITY

Approved

DATE ANSWERED

Sep 18, 2020

EXPERT

Tutor

ANSWER RATING

GET INSTANT HELP/h4>

We have top-notch tutors who can do your essay/homework for you at a reasonable cost and then you can simply use that essay as a template to build your own arguments.

You can also use these solutions:

  • As a reference for in-depth understanding of the subject.
  • As a source of ideas / reasoning for your own research (if properly referenced)
  • For editing and paraphrasing (check your institution's definition of plagiarism and recommended paraphrase).
This we believe is a better way of understanding a problem and makes use of the efficiency of time of the student.

NEW ASSIGNMENT HELP?

Order New Solution. Quick Turnaround

Click on the button below in order to Order for a New, Original and High-Quality Essay Solutions. New orders are original solutions and precise to your writing instruction requirements. Place a New Order using the button below.

WE GUARANTEE, THAT YOUR PAPER WILL BE WRITTEN FROM SCRATCH AND WITHIN YOUR SET DEADLINE.

Order Now