Experiment 10


Refraction of Light (Thin Lenses)



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




   A computer with Internet connection, a calculator, a ruler, a few sheets of paper, and a pencil


(Note: First go to the "Practice Pageunder "Theory," and practice the ray diagrams.  On each diagram, a few guiding points are provided that will help you draw that ray diagram easily.  This must be done before the experiment.)




   The thin lens formula is:

where do (the object distance) and d (the image distance) are both measured from the lens, and  f is the focal length.


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


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



The Ray Diagrams for a Converging lens:


   There are six different cases for image of an object in a converging lens four of which are shown below:



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:



 1) Real  

 2) Inverted

 3) A'B' << AB 

 4) Forms at F  (di = f )




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



 1) Real 

 2) Inverted

 3) A'B' < AB 

 4) f < di < 2f



   Infinite rays emerge from any point of object AB including 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 first and then through the lens, leaves the lens parallel to the main axis, after refraction.  The intersection of these two refracted rays, form A', the image of A.    The third important ray is the one that goes through the center O almost without refraction.  The third ray is shown as a dotted line in the above figure.


   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 Object AB, then image A'B' will result.  In practice, finding the image of one point is enough.  The best point to choose is Point A, the top of the object.  From A', a perpendicular line must then be drawn to the main axis that is named 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.




 1) Real 

 2) Inverted

 3) A'B' >> AB

 4) di  



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.


 Object beyond 2f  (do > 2f )


Image Condition:  1)                 2)                   3)                  4)



Object at 2f  (do = 2f )


Image Condition: 1)                 2)                   3)                  4)


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

Image Condition: 1)                 2)                   3)                  4)

Object almost at F


Image Condition: 1)                 2)                   3)                  4)



Object within f  (do < f )


Image Condition: 1)                 2)                   3)                  4)





Image in Diverging lenses


The image of an object in a diverging (concave) lens is always virtual, upright, smaller than the object, and forms within the focal length on the side that the object is.


Practice on Image in a Diverging Lens




If you have not done so, you may need to add the following Website to your Java exception list: http://surendranath.tripod.com/.   To do this, follow the path (Windows operating system),

Start → All Programs → Java → Configure Java → Security (use High) → Edit Site List … → Add → Type in the site URL (http://surendranath.tripod.com/).

   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 length The value of f can be read from screen top.  As you move one of the F's, 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 has. 


   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.  If the applet says f = 100.0, it means f = 100.0mm.  




Part 1: Convex Lens (Converging Lens):


1) Make sure that the convex lens is exactly at the middle.  Set the F (the Focal Point) at 120units, 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 far left 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


5) Use magnification M to calculate y' (the image height).  This will be the accepted value for y'.  Record all values in Table 1.


6) Keep f = 120mm and repeat the above steps for the following do values:   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 squares to its left.


2) Move the right side F to 8 squares from the lens to 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.

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


4) Using do , f, and y, along with the lens and magnification formulas, calculate the accepted values for di and y'.


5) Finally calculate the necessary % errors.


6) Keeping the same object height of y = 80.0mm, and f = -160.0mm, repeat the above steps for the following do values:  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.



       Given and Measured:


Case do (mm) f (mm) Measrd




Magn. |M Obj. Size




Image Size (y')



Image Size(y')










Convex Lens:
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        




 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.