EXPERIMENT 12 CD DIFFRACTION

EXPERIMENT 12 CD DIFFRACTION

 

 

 

Introduction: For this experiment we will be investigating the difraction by reflection. this will be done by striking a laser beam in a cd. this cd contains grooves that act as a reflective grating. with this experiment we well be investigating the equation for diffraction. 

 

 

 

Here are some pictures of our set up.

 

brief description we flashed our laser trhough a hole in our paper and then hitting the cd. the laser then was reflected but it was refracted by the groves in the cd. the distance between our two shining points is the distance between our two first ordere maximas.

 

 

in ordere to proceed in fiding the distance betwween each grove we firs require to know the wavelenght of the laser. this lead to another experiment using a diffrating grating. the top of the  picture is the calculation in finding the wavelenght. once we found our wavelenght we then proceed with our calculation in finding the distance between the grooves of the cd.

 

here is the calculation for the uncertainity of the distance between the grooves.

 

pik

 

 

 

 

in conlusion it was found that with the aid of  the grating slit difraction equation we were able to find the distance between each grove. Our final answer for such distance was to be 1565,31 +- 61.19 nano meters This result fell between the manufaturers standard value wich is about 1600 nm. this proves the validitiy of such equation. 

.

 

 



 

 

 

 

 

  Experiment 9

Lenses

 

 

In this experiment we will be investigating the properties of lenses and the lens maker equation.the purpose of this lab is to identify the difference in images when an object is placed in different points from the lenses. here is one of the most useful forms of the lens maker equation

You can see the setup of this experiment in the lab manual.

                                             

                                                  here are some images of the experiment

                                                      

=.-

After proceeding in finding the focus of our lenses, we found this focus to be 5.2 +,- .5cm. here is the table of our measurements after moving our object in different positions from the focus in order to obtain a different image distance and a different image height. 

                                                   

                               As you can see we gave a reasonable error to our measurement

Here is the procedure we fallow when obtaining the magnification of the lensesfor one of the measurements with its proper procedure in obtaining Uncertainty.

 

 

 

 

 

Answers to question 4.

1.- when the object is less than the focus, the image does not appear in the white board it only appears in the lenses. This is due to the fact that the image is virtual. 
                                                  here is a picture of a virtual image

 

 

 

Graph of di(distance of the image) vs do(distance of the object).

do	di
26	6.28
20	7
15.6	7.4
10.4	9.8
7.8	16

 

 

 

 

Here is the graph of the inverse of di and do

 

slope = .9478    y intercept = .18

 

The y intercept represents :1/focal

 

 in overall we observed that the equation of lensmaker is the main ruler when dealing with objects reflected in lenses. We also observed that images height changes whenever we move the position of our object. When we know our focal we are able to know when does an image stops from being real and turns in to virtual and vise versa.  

 

Experiment 8 concave and convex mirrors

The purpose of this exploration is to gain knowledge on the reflecting images on concave and convex mirrors. We will explore the different images portrayed on the mirror whenever we move towards and away from the focal.

 

Convex mirror:

Answers to questions a-c

a)      In a convex mirror the image seems smaller

b)      The image is upright

c)      The image is behind the mirror

 

2) When the object is moved towards the mirror nothing seems to happens. The image does not depend on the distance of the image

3) When the object is moved away from the mirror the image is still in the same position. The image does not depend on the distance.

 

Here is the picture of the image and the calculated magnification.

 

9) The magnification value seems to be consistent with the sketch.

 

 

 Concave mirror

 

Answers to questions a-c

a)      In a concave mirror the image varies with position. The image is inverted and and as the object gets closer it increases in size but inverted until the focal point is approach then the image disappears, in front of the focal the image appears large but behind the mirror.

b)       The image is inverted if it is behind the focal point

c)      The image is in front of the mirror

 

2) The image increases size but inverted as the object gets closer

3) When the object is moved away the image seems to get smaller.

Here is the picture of the image and the calculated magnification.



 

9) The magnification value seems to be consistent with the sketch.

In conclusion we found out the image of a convex mirror does not depend on the distance of the object. This image is always virtual and erect. When we are using a concave mirror the image does depend on the distance on the object. As we get close to the mirror the image is always inverted and it gets bigger. When we hit the focal point the image disappears. As we pass the focal point the image is virtual and erect, and the magnification is bigger.

Experiment 7 Reflection and Refraction

Experiment 7

 

in this experiment we will begin to study optics first by beginning with reflection and refraction. In this experiment we should be able to understand Snells law and the variables that govern such equation.

Answers to questions pre starting lab.

 

a)      The angle will be 0

b)      The angle of refraction will also be 0

c)      As the light leaves the plastic, this light will diffuse in air due to the fact that the curve surface will cause the light to bend since it is not flat.

d)     The experiment will be a situation in which the light travels from an material of higher density to a material of lower density.

 

Answer to question #2

 

The light ray does behave as predicted when the angle is 0

 

here is the first measurements

liight hiting the acrylic straight edge first
incidence angle(θ1)	refraction angle(θ2)	sin (θ1)	(θ2)
10	7	0.1736	0.121
15	12	0.258	0.2
20	14	0.342	0.241
30	20	0.5	0.342
35	23	0.573	0.39
40	25	0.64	0.422
45	30	0.7	0.5
50	32	0.766	0.529
60	36	0.866	0.587
70	41	0.939	0.656

here is the graph of the sin(theta1)/sin(theta2) values with its proper linear function

Answers to questions 6,7,
answer to #6.
The slope represents the index of refraction of the air over the index of the acrilyc.

answer to #7

  Answers to questions pre starting part 2

a) The angle will be 0

b) The angle of refraction will also be 0

c) As the light leaves the plastic, this light will diffuse in air due to the fact that the curve surface will cause the light to bend since it is not flat.

d) The experiment will be a situation in which the light travels from an material of lower density to a material of higher density.

 

here is the second measurments

 

light hitting the acrylic with the circular edge first
incidence angle(θ1)	refraction angle(θ2)	sin (θ1)	(θ2)
0	0	0	0
5	7	0.087	0.121
10	15	0.173	0.258
15	24	0.258	0.406
20	35	0.34	0.57
30	53	0.5	0.79
40	75	0.64	0.96
44	90	0.69	1

 here is the graph of   sin(theta1)/sin(theta2) the values  and the linear equation of the slope

Answers to questions  6,7,
 answer to #6. 
the slope represents the index of refraction of the acrylic over the index of air

answer to #7                                  
                                  here is a brief explanation of how to come about to snells law.

 

Answers to questions 10,11,12.

10) 

For the second part we were not able to complete the 10 trials due to the fact that as we encounter the 45 degree angle the diffracted light disappeared This angle is called the critical angle.

11)
The slope of this second graph is 1.45 which is the ratio of the reflecting index of the acrylic over air

12)
Here is the equation

y = 1.4754x + 0.0199

 

In conclusion we found out how the ray of a light wave behaves when there is a change in media (medium). This light ray refracts when entering a different density medium.this refraction continues as the angle of incidence in creases. there is an angle when the angle of refraction will be just reflected and this angle is called the critical angle Also with our graphs we were able to obtain the index of refraction of the acrilyc and this helped us in understanding  Snells law.