Experiment 9.2

Refraction of Light ( Thin Lenses)

Objective:

 

The objective is to verify the thin-lens formula by forming the image of an object in a converging (convex) lens.

 

Equipment:

 

A lens holder, a converging (convex) lens, an optical bench, a light-bulb socket, a low-watt light bulb, a rectangular target, a target holder, and a tape measure

 

Theory:

 

   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, any virtual quantity is negative, and any real quantity 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:

 

 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 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 objectIf 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.

 

 

 Image:

 1) Real 

 2) Inverted

 3) A'B' >> AB

 4) di 

 

 

The practical case is shown below:

 

 

 

Case VI: Object within f  (do < f )

 

 

Image:

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

 

 

Procedure:

 

1.                  Turn a dim light bulb on that is already placed in a holder and mounted on an optical bench.  The bench must be at one end of the room.  Turn the room lights off.

 

2.                  Obtain a converging lens and hold it near the other end of the room such that light from the bulb can pass through it.

 

3.                  By moving the lens back and forth toward and away from the wall, try to form a clear image of the light bulb on the wall.  Do you expect to see the image upright or inverted?

 

4.                  Since the light is coming from a relatively far object, the rays are nearly parallel and the image will form almost at the focal point of the lens.  It is possible to measure the distance from the wall to the lens' center and accept it as an approximation to f, the focal length.  Measure this distance just for curiosity.  This is not the best way to measure f unless parallel rays of sunlight form a tiny spot of light on a piece of paper.

 

5.                  Steps 1 through 4 may be performed without a light bulb if the room has windows.  The image of the window and the outside scene as seen through it can be formed on a vertically held cardboard or a wall.  The room lights must be off.

 

6.                  To measure a more precise value for f , case III of the image in a converging lens may be used.  Case III is the case where the object is at 2f and the image forms at 2f as well.  In other words, object and image are equidistant from the lens in this case.

 

7.                  On an optical bench, mount a low-watt light bulb and a converging lens on one side of it as close to it as possible.  Then place a screen (a target in a target holder) on the other side of the bulb, also as close to the bulb as possible.  Light from the bulb must go through the lens and reach the screen (target) as shown below:

 

 

8.      Simultaneously move the bulb and the target away from the lens such that at any state they are equidistant from the lens.   Doing this, there comes a distance at which an inverted and clear image of the bulb forms on the target.  Of course, the image will be equal to the object (the light bulb).  According to Case III, both the object and its image are at 2f .  Make sure that the image distance is equal to the object distance while the image is at its best.  Measure this distance and calculate f from it.

 

  9.     Once f is determined, try the following 3 cases:

 

Place the object (the light bulb) at

 

a)      do = 2f + 12.0cm,

b)      do = 2f - 3.0cm, and

c)      do = f + 0.80cm.

 

In each case search for the image position di  by adjusting the target position back and forth.  Measure di in each case and record its value.   If the image is big, you may have to form the image on the walls of the lab or room you are in.  The room must be kept relatively dark.

 

10.     In each case use the f found in Step 8 along with the given di  in Step 9 to calculate the accepted value for di using the thin-lens formula.

 

11.     For every measured di , there is an accepted value found in step 10.  Calculate a % error for each to see how well the thin-lens formula verifies each case.

 

Data:

Given:

                       

Object distance do  as selected by students in your group or   suggested as

                       

a)      do = 2f + 12.0 cm,

b)      do = 2f - 3.0 cm, and

c)      do = f + 0.80 cm.

 

Measured:

           

The value of di for each given do 

 

 Calculations:

 

Show your calculations.

 

Comparison of the Results:

 

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

                       

Conclusion:  To be explained by students.

 

Discussion:   To be explained by students.