Invese Squae Law and Polaization Objectives: To show that light intensity is invesely popotional to the squae of the distance fom a point light souce and to show that the intensity of the light tansmitted though two polaizes depends on the squae of the cosine of the angle between the axes of the two polaizes. Appaatus: 1 Pasco Optics Bench (OS-8518) Point light souces (OS-8517) 1 Photomete with filte set (OS-850) Polaizes (OS-850) Explanation: Invese Squae Law: The light fom a point light souce speads out unifomly in all diections. The intensity at a given distance,, fom the light will be equal to the powe output of the light divided by the suface aea of the sphee though which the light has spead. Since the aea of the 1 sphee vaies as the squae of its adius,, the intensity will dop off as. In geneal, the intensity of the point light souce at any distance,, is given by constant I = (1) Thus, the atio of the intensity (I) of the light at a position () as compaed to the efeence intensity (I 0 ) measued at a position ( 0 ) is given by I I 0 = 0 Polaization: A polaize only allows light which is vibating in a paticula plane to pass though it. This plane foms the axis of polaization. Unpolaized light vibates in all planes. Thus if unpolaized light is incident upon an ideal polaize, only half will be tansmitted though the polaize, and the tansmitted light is polaized in one plane. (In eality, no polaize is ideal and less than half the light is tansmitted.) If this polaized light is incident upon a second polaize, the axis of which is oiented pependicula to the plane of polaization of the incident light, no light will be tansmitted though the second polaize as in the figue below. () 1 Polaization
Howeve, if the second polaize is oiented at an angle that is not pependicula to the fist polaize, thee will be some component of the electic field of the polaized light that lies in the same diection as the axis of the second polaize, and thus some light will be tansmitted though the second polaize as in the figue on the next page. The component, E, of the polaized electic field, E 0, is E = E 0 cosφ. Since the intensity of the light vaies as the squae of the electic field, the tansmitted light intensity is given by I= I 0 cos φ (3) whee I 0 is the incident light intensity and φ is the angle between the axis of polaization of the incident light and the polaize. How to Use the Photomete: To detemine if the two sides of the photomete ae illuminated by light of equal intensity, look down into the conical eyepiece of the photomete. The cone is designed to cast a shadow on the inne pats of the photomete to allow a bette view. Photomete Do not put you eye diectly on the eyepiece. Keep you head at a distance which allows you to comfotably focus on the oange indicato. By looking though the eyepiece, you can see if the disks, each eceiving light fom an opposite side of the photomete, ae equal in intensity. Although it is difficult fo the eye to detemine elative intensities, it can detect equal intensities quite accuately Polaization
Setup fo Pat 1: Invese Squae Law 1. Place the photomete at the 70 cm mak on the optics bench.. Place a point light souce at 40 cm. This is light souce 1. Put a neutal density filte on the side of the photomete that is opposite the point souce. Place the othe light souce on the same side of the bench that has the neutal density filte. This is light souce. 3. Adjust the neutal density filte fo 100% tansmittance. WHEN MOUNTING AND UNMOUNTING THE LIGHT SOURCE, DO NOT REMOVE THE SQUARE NUTS. Pocedue Pat 1: Invese Squae Law NOTE: You may want to cove the cossed-aow object on each light souce to educe the excess light in the oom. The lights in the oom need to off o vey dim. Remembe to include uncetainty of you measuements and to popagate eo. 1. Look into the photomete and move the light souce (on the filte side) to a position that gives equal intensities. The light souce will emain at this position fo the est of the expeiment and will seve as the efeence intensity I 0. Recod the positions of the photomete and the light souce 1 (opposite the filte side of the photomete) in Table 1. The position of the efeence light is not needed.. Rotate the neutal density filte to 75% tansmittance. Move the point light souce 1 to the position whee the intensities ae once again the same when viewed in the photomete. Recod this new position of the light souce in Table 1. 3. Repeat the last step fo 50% and 5% tansmittance. 4. Without moving the photomete, epeat the above steps fo thee moe tials. 3 Polaization
Analysis: 1. Using the measued positions in Table 1, calculate the distances of the point souce fom the photomete and ecod in Table 1.. Fo each of the diffeent positions, calculate the intensity in tems of I 0. Use 0 as the initial distance of the point souce (100%) and as the distance at the given intensity. Note that the intensity is calculated in tems of the initial intensity I 0. Recod you answes in Table 1. 3. Calculate the pecent diffeence between the calculated intensities and thei coesponding expected values. Recod in Table 1. Questions: 1. Based on you esults, how does the intensity of light vay with distance? (Explain in thee o moe sentences.) Setup fo Pat : Polaization 1. Use the same geneal set up as in Pat 1.. Snap one polaize onto each side of the accessoy holde. Befoe beginning the expeiment, check the angle calibation on the polaizes in the following way: On the side of the accessoy holde that has the label, set the angle to 90 degees. Look though both polaizes at a bight light and otate the othe polaize until the tansmitted light is at the minimum. Now the polaizes ae cossed at 90 degees. Rotate the label side polaize back to zeo degees. 3. Now the two polaizes ae aligned fo maximum tansmission. 4. Thoughout the expeiment, only otate the label-side polaize. 5. Place the polaize accessoy holde (with polaizes) on the bench between the light souce and the photomete on the side opposite the neutal density filte. The label side of the polaize holde should face away fom the photomete. The polaize holde should be close to the photomete so only polaized light will ente that side of the photomete. Pocedue pat : Polaization NOTE: You may want to cove the cossed-aow object on each light souce to educe the excess light in the oom. The lights in the oom need to off o vey dim. 4 Polaization
Remembe to include uncetainty of you measuements and to popagate eo. 1. Set the neutal density filte fo 100% tansmission.. While looking into the photomete s conical eyepiece, adjust the position(s) of the light souce(s) until the two sides of the oange indicato have equal intensity. 3. Set the neutal density filte fo 75% tansmission. 4. While looking into the photomete s conical eyepiece, otate the label-side polaize until the two sides once again have equal intensity. Recod the angle in Table. Rotate the polaize back to zeo and epeat the measuement thee moe times. 5. Repeat the pevious step fo 50% and 5% tansmission. Analysis 1. Fo each of the neutal density filte settings, calculate the aveage of the thee tials and ecod the aveage angle in Table.. To calculate the pedicted pecentage tansmittance fo each case, calculate the squae of the cosine of each aveage angle and ecod in Table. 3. Calculate the pecent diffeence between the pecentage tansmittance and the pedicted value fo each case and ecod in Table. Questions: 1. Since an ideal polaize does not exist, less than half the light is tansmitted. How does this impact you esults?. Did you veify Eq. (3)? 3. How could you detemine the polaization axis of a pai of polaized sunglasses? 5 Polaization
Name: Patne: Date: Invese Squae Law and Polaization Photomete Position = Table 1: Invese Squae Law Light Souce Intensity 100% 75% 50% 5% Tail 1 Position 1 Tail Tail 3 Tail 4 Aveage Position of Point Souce Distance fom Photomete Calculated Intensity (% of I 0 ) % diffeence Table : Polaization % Tansmittance 75% 50% 5% Tial 1 Tial Tial 3 Aveage angle φ cos φ % diffeence Wok Safely
Wok Safely