Experiment 10

 

Refraction of Light ( Thin Lenses)

Objective:

 

To verify the thin-lens formula by forming the image of an object in both converging (convex) and diverging (concave) lenses.

 

Equipment:

 

A computer with Internet connection, a calculator (The built-in calculator of the computer may be used), a ruler, a few sheets of paper, and a pencil

 

(Note:  Before the experiment, get a ruler and go to the Practice Page under "Theory," and practice the ray diagrams.  On each diagram, a few points are provided that will guide you as how to draw that ray diagram.  This must be done preferably before the experiment.  It is o.k. to to it after the experiment, if you try hard and do not understand it.)

 

Theory:

 

The thin lens formula is:

 

where  do is the object distance from the lens,  di  is the image distance from the lens, and  is the focal length of the lens.

 

Assigning a (-) sign to di indicates advance knowledge of a virtual image and obtaining a (-) value for the image through solving this equation for di indicates a virtual image as well.

 

For a diverging lens, ( f ), the focal length should be given a  ( - ) sign when using the above equation, because for a diverging lens, even the focal point is virtual.  In general, anything virtual is negative, and anything real is positive.

 

Magnification:  Magnification is the ratio of the image size to the object size.  The absolute value of magnification is given by the following formula:

 

 

Ray Diagrams for a Converging lens:

 

There are six different cases for image of an object in a converging lens.

 

 

Case ( I ): Object at infinity ( do >> 2f )

 

Rays coming from very far away are practically parallel.  If such rays are also parallel to the main axis of the lens, the image forms at ( F ), the focal point of the lens, as shown:

 

 Image:

 1) Real  

 2) Inverted

 3) A'B' << AB 

 4) Forms at F (di = f)

 

 

 

Case ( II ): Object beyond 2f ( do > 2f )

 

 Image:

 1) Real 

 2) Inverted

 3) A'B' < AB 

 4) f < di < 2f

 

 

 

Infinite rays emerge from object AB, even from point A of it..  Two rays are selected.  Important Ray 1 that travels parallel to the main axis passes through F, after refraction.   Important Ray 2 that goes through F before going through the lens, leaves the lens parallel to the main axis, after refraction.  The intersection of these two rays, form A', the image of A.    The third important ray is the one that passes through the center ( O ) without refraction.  The third ray is shown as a dotted line.

 

Any two of these three important rays may be used to form the image of A, the top of the object.

If the same procedure is repeated for all points of the object (AB), then image A'B' will result.

In practice, only finding the image of A, the top of the object is enough.  From A', a perpendicular line to the main axis, gives B', the image of B.

 

For other cases, apply a similar method to form A'B', the image of AB.

 

 

Case ( V ): Object at F ( do  = f ) In practice, object is placed slightly passed f

 

 

 Image:

 1) Real 

 2) Inverted

 3) A'B' < AB

 4) f < di < 2f

 

 

The practical case is shown below:

 

 

 

Case ( VI ): Object within f ( do < f )

 

 

1) Virtual               

2) Upright

 3) A'B' > AB     

4) Image forms on the same side the object is.

 

 

 

 

 

Practice Page:  Use two out of three important rays emerging from A to form its image (A') and complete each of the following ray diagrams:

Note that A' is found by the intersection of rays refracted through the lens.

Also state the image conditions.

 

II)  Object beyond 2f ( do > 2f )

 

Image Condition:  1)                        2)                        3)                    4)

 

 

III)  Object at 2f ( do = 2f )

 

Image Condition:  1)                        2)                        3)                    4)

 

IV ) Object between f and 2f ( f < do < 2f )

Image Condition:  1)                        2)                        3)                    4)

V ) Object almost at F

 

Image Condition:  1)                        2)                        3)                    4)

 

 

VI ) Object within f ( do < f )

 

Image Condition:  1)                        2)                        3)                    4)

 

 

 

 

Image in Diverging lenses

 

A diverging (concave) lens forms an image of an object that is always virtual, upright, smaller than the object, and at the same side as the object is.

 

Practice on Image in a Diverging Lens

 

 

Procedure:

 

Click on the following link:  http://surendranath.tripod.com/Applets.html .   Click on the "Applet Menu", then on "Optics", and then on "Spherical Mirrors and Lenses."  The applet asks you to click on it to start.  Click on the applet.  The screen for the experiment (or the Optical Bench) appears.  On the top left dropdown menu, click on "Convex Lens" if a convex lens is not already on the bench.  The convex lens allows you select a focal length for it.  On each side of the lens, on the main axis, you should see two yellow dots. Those are f and 2f distances from the lens.  Placing the mouse on F of one side and moving it left or right changes the focal lengthThe value of f can be read from screen top.  As you move one of the Fs, you will see that the convexity of the lens changes.  The more convex a lens gets, the shorter its focal length becomes and a stronger convergence power it gains. 

 

If the lens is placed exactly at the middle, there will be 30 squares on each side of it. Each square is 20units X 20 units.  The minimum focal length, f, this applet allows is 100 units.  We are going to think of (mm) as the unit of length in this experiment.  If the applet says f = 100.0, it means f = 100.0mm.  

 

Experiment:

 

Part 1, Convex Lens (Converging Lens):

  1. Make sure that the convex lens is exactly in the middle.  Set the F (the Focal Point) at120units, or let f = 120mm.  Read the top of the applet to make sure it reads f = 120mm. 

  2. Place the mouse at the tip of the object (the red arrow) and make its height equal to 4 squares or 80.0mm.  Also move it to the left far enough at do = 400.0mm.  Check the values of y ( the object height), and do on the top to assure their correctness.

  3. Measure (Estimate) the position of di on the applet and record your estimate in Table 1.   Estimate the height of the image, y',  on the applet and record your estimate in Table 1.  These will be your measured values.   In estimation, each square has a length of 20.0mm.

  4. Use the given values of  do and f to calculate the expected di.   This will be the accepted value for di.   Use this di and do to find the magnification, M.   Then use the magnification, M to calculate y' (the image height).  This will be the accepted value for y'.  Record all values in Table 1.

  5. Keeping f = 120mm, repeat the above steps for the following do values:  do = 240.0mm, 170.0mm, 141.0mm (with y=40mm for this case to keep the image in range), 121.0mm, 119.0mm (with moving the lens to the far right for this case), and 70.0mm (you may want to move the lens to the middle and change y to 80.0mm again).  Note:  In cases that the image goes out of screen, just calculate the accepted value for di and y', and leave the space for the measured values blank and do not calculate their respective %errors.

 

Part 2, Concave Lens (Diverging Lens):

  1. Change the lens to a concave one by clicking on the "Concave Lens" in the dropdown window.  A concave lens appears.  With the mouse at its center, place its center exactly at the16th square from the right side of the screen.  The screen is 60 squares wide.  Now, there must be exactly 16 squares to the right of the lens and 44 to its left.

  2. Move the right side focal point F to 8 squares from the lens ( f =160.0mm).  This places the virtual right side 2F at 320.0mm from the lens or at the rightmost edge of the matrix.  Make sure the applet reads the same.  Of course, you know that the focal point and focal length of a concave lens are both virtual.  The f to be used in calculations is actually f = -160.0mm.

  3. Keeping the object height y = 80.0mm, move the object to 3 squares from the lens at do = 60.0mm.  Double-check you readings for f, do , and y.

  4. Again, measure (estimate) di and y'.  These will be your measured values.

  5. Using do , f, and y, along with the Lens and Magnification formulas, calculate the accepted values for di and y'.

  6. Finally calculate the necessary %errors.

  7. Keeping the same object height of y = 80.0mm, and f = -160.0mm, repeat the above steps for the following do values:  do = 120.0mm, 230.0mm, 320.0mm, 500.0mm, and 720.0mm (you may have to move the lens a few squares to the right for this case).  Do all calculations for each case and record the values in Table 1.

 

Data:

Given and Measured:

 

Case Object

 Distance

do

(mm)  

Focal

 Length

f

(mm) 

  Measured

Image Dist.

 di

(mm)

 Accepted

Image Dist.

di

(mm)

Absolute Value of Magnifi -cation

M

Object

Height

y

(mm)

Measured

Image Size

y'

(mm)

Accepted

Image Size

y'

(mm)

%

Error

on

di

 

%

Error

on

y'

 

Convex Lens:       80.0        
1 400.0 +120.       80.0        
2 240.0 +120       80.0        
3 170.0 +120.       80.0        
4 141.0 +120.       40.0        
5 121.0 +120.       40.0        
6 119.0 +120.       40.0        
7 70.0 +120.       80.0        
Concave Lens: Virtual Image Position ( di ) is Negative                    
8 120.0 -160.0 -            -              80.0        
9 230.0 -160.0 -            -              80.0        
10 320.0 -160.0 -            -              80.0        
11 500.0 -160.0 -            -              80.0        
12 720.0 -160.0 -            -              80.0        
                     

 

Calculations:

 

 Show 2 full calculations for each lens.

 

Comparison of the Results:

 

Calculate a %error on di and y' for each case using the usual %error formula.

                       

Conclusion:   To be explained by students.

 

Discussion:    To be explained by students.