### Experiment 7

#### Refraction of Light ( Thin Lenses)

Objective:

The objective is to verify the formula of thin lenses 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 is the object distance from the lens,  d  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.

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 leaves the lens parallel to the mail 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) di >>f

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

Practice on Image in a Diverging Lens

Procedure:

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.

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

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?

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.

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.

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.

On an optical bench, mount a light bulb (a low watt bulb does not bother your eyes) and a converging lens on one side of the lit bulb and as close to it as possible.  Also place a target placed in a target holder on the other side of the bulb also as close to it as possible.  The bulb sends light through the lens and the refracted light shines on the target.  See Figure below:

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 clearest state possible.  Measure this distance and then calculate ( f ) from it.

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.

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 mirror formula.

For every  di measured, there is an accepted value found in step 10.  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 + 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:

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