Experiment 8.2

Reflection of Light (Spherical Mirrors)

Objective:

 

The objective is to verify the formula of spherical mirrors by forming the image of an object in a converging (concave) mirror.

 

Equipment:

 

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

 

Theory:

 

The formula for spherical mirrors is

where  do is the object distance from the mirror,  d  is the image distance from the mirror, and  f   is the focal length of the mirror.    f = ½ R with R being the radius of the mirror.  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 mirror, ( f ), the focal length should be given a  ( - ) sign when using the above equation, because for a diverging mirror, even the focal point is virtual.  In general, anything virtual is negative, and anything real is positive.

 

Ray Diagrams for a Converging Mirror:

 

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

 

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 mirror, the image forms at ( F ), the focal point of the mirror, as shown:

 

Image:

1)Real                    2) Inverted

3) A'B'<<AB         4) 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 reflection.   Important Ray 2 that goes through F leaves the mirror parallel to the mail axis, after reflection.  The intersection of these two rays, form A’, the image of A.   A third ray could have been drawn whose reflection would pass through A’ as well.  It is shown as a dotted line in the diagram.  This third important ray is the one that passes through the center ( C ) and it reflects back on itself.  For a ray passing through C, the angle of incidence is zero and so is the angle of reflection; therefore, it reflects back on itself. 

 

Case ( III ): Object at 2f ( do = 2f )

Image:

1)Real                    2) Inverted

3) A'B' = AB         4) di = 2f

 

 

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

Image:

1)Real                    2) Inverted

3) A'B' > AB         4) di > 2f

 

 

Case ( V ): Object at F( do  = f )

Image:

1)Real                    2) Inverted

3) A'B' >> AB         4)  di → ∞

 

 

Case ( V' ): Object slightly beyond F( do ~ f )

Image:

1)Real        

2) Inverted

3) A'B' >> AB

 4)  di → ∞

 

 

 

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

 

Image:

1)Virtual              2) Upright

3) A'B' >AB    

4)  di behind the mirror

 

 

 

 

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 reflected from the mirror.  Also state the image conditions.

 

 

 

 

Image in Diverging Mirrors

 

A diverging (convex) mirror forms an image of an object that is always virtual, upright, smaller than the object, and behind the mirror.

 A ray diagram is shown below:

Image:

1)Virtual              2) Upright

3) A'B' <AB    

4)  di behind the mirror

 

 

Practice on Image in a Diverging Mirror

 

Complete the ray diagram shown below:

Procedure:

 

1.                  Turn a dim light bulb on (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 mirror and hold it near the other end of the room such that its shiny (reflecting) side faces the dim bulb.

 

3.                  Hold a sheet of paper or cardboard in a vertical position such that the reflected light from the mirror projects on it.  Make sure the cardboard or paper does not block light from reaching the mirror.

 

4.                  By moving the vertical cardboard back and forth in front of the mirror, try to form a clear image of the light bulb on it.  Do you expect to see the image upright or inverted?

 

5.                  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 mirror.  It is possible to measure the distance from the cardboard to the mirror’s vertex 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 ).

 

6.                  Steps 1 through 5 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 paper.  The room lights must be off.

 

7.                  To measure a more precise value for ( f ), case III of the image in a converging mirror 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 mirror in this case.

 

8.                  On an optical bench, mount a light bulb (a low watt bulb does not bother your eyes) and a converging mirror facing the lit bulb.  Also hold a target placed in a target holder just beside the bulb so that it faces the mirror as well.  The bulb sends light to the mirror and the reflected light shines on the target.  See Figure below:

9.        Slowly move the mirror away from the bulb-and-target pair.  The mirror must be slightly turned such that the reflected light shines on the target.  Frequent adjustments are necessary.  As the mirror is moved away from the pair, there comes a distance at which and inverted and clear image of the bulb forms just beside the bulb but 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 ).  Measure this distance and then calculate ( f ) from it.

 

10.       Once ( f ) is determined, try the following 3 cases:

 

11.       Place the object (the light bulb) at

 

a)      do = 2f + 8.0cm,

b)      do = 2f - 3.0cm, and

c)      do = f + 3.0mm.

 

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.  In each case use the ( f ) found in step 9 along with the given ( di ) in step 10 to calculate the expected ( accepted ) value for ( di ) using the mirror formula:

12.       For every ( di ) measured, there is an accepted value found in step 11.  Calculate a %error for each to see how well the mirror formula verifies each case.

 

Data:

 

        Given:

 

                  Object distance ( d ) as selected by students in your group or suggested as

                       

a)      do = 2f + 8.0cm,

b)      do = 2f - 3.0cm, and

c)      do = f + 3.0mm.

 

Measured:

           

A value of ( di ) for each given( do )

 

Calculations:

 

Use the mirror’s formula:

 

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