Determination Of Refractive Indices For A Prism Material And A Given Transparent Liquid 51606b

Experiment :

Study of the index of Refractive-Line of Demention

Object

:

Determination of refractive indices for a prism material and a given transparent liquid

Apparatus

:

Spectrometer, Na light source, Prism with one face grounded, glass slide.

Theory

:

If the incident angle of a light ray on DF surface, coming from a point P, is greater than the critical angle C (see figure 1(a)), it will subject to a total internal reflection, and travel along QRS direction .rays from P which incident within FQ range are refracted while those incident within QD range are internally reflected a the DF surface. Therefore, a margin may be observed from SR direction where the intensity varies from brightness (RD region) to darkness (ER region) to darkness (ER region). The two regions are separated by a line of demarcation and this position could be observed through the telescope of the spectrometer. When the surface DF is covered with a film of liquid (e.g. water) a similar demarcation line also cold be observed.

From Snell’s Law, At Q ; At R ;

 sin C  1  (1)  sin   sin   (2)

Here,  and  1 are the indices of refraction for the prism substance and the medium next to the DF surface, respectively.  and  denote the angles of incidences and emergence at the ED surface, Respectively. From GQR triangle (Figure 1),

C  A  (3); C  A.

From DQR triangle (Figure 2),

C  A   (4); C  A.

(1),(2) and (3) result in 1  sin A( 2  sin 2  )

1

2

 cos A sin   (5)

For the situation as in figure 1(a) with air on surface DF, we have  1 =1. Then the index of refraction for the prism substance is, 1

 1  cos A sin   2  2 2    (6)   sin    sin A   

For the situation as in figure 1(b) with the liquid next to the surface DF, the refractive index of the liquid  1 can be determined as follows.

From (1),(2) and (4) we have, 1   sin( A   )

1  sin A(  2  sin 2  )

1

2

 cos A sin   (7)

By substituting  obtained from (6) and the corresponding  Angle measured when there is liquid next to surface DF, in equation (7) we get refractive index of liquid (  1 ) Determination of the angle  when air is present next to the surface DF: When the DF surface has air next to it, one can observe the demarcation line through the DE surface, which has an angle of emergence  at R (see figure 2). A similar observation can be made through the DF surface

with the same  at Q as shown in the figure 2. If 0is the angle between two demarcation lines, one can write from QPRS rectangle that

Figure 2

(180  A)  2  (360   )  360,

These yields 

  A  180 2

 (8)

Determination of the angle  when the liquid is present next to the surface DF: When the DF surface is touched by a liquid, one can observe a situation as illustrated in figure 1(b). Do the following steps to determine  in this case. First, place the Na lamp at position (2) and recorded the reading (3) which is the reflection of (2) as shown in the figure 1(b). Now, without changing the prism or collimator positions, place the Na lamp to illuminate the demarcation line. Remove the prism to record the reading (4) [direct reading of (2)]. From these data, one can determine  as 

(3)  (2)  (3)  (1)   (9) 2

Procedure

:

a. The spectrometer was adjusted for parallel rays using Schuster’s method. b. The prism was Leveled table and obtain the prism angle A c. The slit was Remove and the prism was placed such that rough surface is facing the collimator. d. Then the telescope was rotated to observe demarcation lines from other two sides for glass-air case. e. A little amount of liquid was Apply of which the refractive index is to be determined (water) on one plane surface of the prism using a glass plate and demarcation was observed line for glass-liquid case. f.  was determined for air from (7) and for liquid form (8) to calculate the indices of refraction for the prism substance and the liquid, respectively.

Calculations: According to first steps; A  6008

  1   2   26621  9525   17056   A  180 

2 17056   6008  180  2   2532  1

 1  cos A sin   2  2 2      sin   sin A    1

 1  cos 6008 sin 2532  2  2 2      sin 2532 sin 6008      1.4655

Direct reading of (2) is Reading of (1) is Reading of (3) is 

= = =

17406 23248 26405

(3)  (2)  (3)  (1) 2

26405  (17406)  26405  23248 2   1343



1  sin A(  2  sin 2  )

1

2

 cos A sin 

1  sin 6008(1.46552  sin 2 1343) 1  1.3721

1

2

 cos 6008 sin 1343

Error Calculations: T1  T2 2 ln A  ln T1  T2  ln 2 A

A A

 T1  T2



2  T1  T2 

2 Least measurement for spectrometer = 0.5  1.4544 10- 4 rad T  T2 A  1 2 2  1.4544 10- 4 A   1.4544 10- 4 rad 2

  1   2   1   2   2  1.4544 10 -4 rad  2.9088  10  4 rad  

  A  180 2

ln   ln   A  180  ln 2

 

   A  180



 

2    A  180

  A

2

2   4.3632  10  4 rad  1  cos A sin    2   sin   sin A    2

  

1 2

 1  cos A sin   2 let k     sin  sin A   then 2

k

1

2

1 ln k 2  1 k  1 k     2 k  2 2

ln  

 1  cos A sin   sin A(cos A cos  sin  sin AA)  1  cos A sin   cos AA  k  2    2 sin  cos sin A sin 2 A      1  cos 6008 sin 2532    2  sin 6008      sin 6008(cos 6008 cos 2532  sin 2532 sin 6008A)  1  cos 6008 sin 2532 cos 6008A  k    sin 2 6008    2 sin 2532 cos 2532      4 and   4.3632  10 rad

k  2  1.4007  4.6439  10 4  3.3937  10 4  k  1.6403 103

 

1 k 2

 

1 (1.6403  10 3 )  5.5964  10 5 2 1.4655

For Part II



(3)  (2)  (3)  (1) 2

ln   ln

 

(3)  (2)  (3)  (1) 2

 (3)   (2)   (3)   (1)

2   2 1.4544 10 -4 rad  2.908810  4 rad

1  sin A(  2  sin 2  )



1

2

 cos A sin 

ln 1  ln sin A(  2  sin 2  )

 sin A(  2  sin 2  )

1

2

 cos A sin 



1

2  cos A sin  1  1 1 sin A(  2  sin 2  ) 2  cos A sin 

1   sin A(  2  sin 2  )

1

2

 cos A sin 

1 1 2 1   2 2 2 sin A  (   sin  ) 2 (2  2 sin  cos)  (   sin  ) 2 cos AA 1   2   cos ACos  sin   sin AA  1 1  2 2 5 2 sin 6008  2 (1.4655  sin 2532 ) (2 1.4655  5.5964 10  1  1   2 sin 2532  cos 2532   2.908810  4 )  (1.46552  sin 2 2532 ) 2 cos 6008 1.4544 10 -4  cos 6008Cos 2532   2.9088 10  4  sin 2532  sin 60081.4544 10 - 4  

1   5.4269 10 5  1.0146 10  4  1.3072 10  4  5.4355 10 5  1  1.2356 10  4

       

Conclusion: Reflective indices of prism according to experiment Standard value for reflective indices of glass Reflective indices of liquid according to experiment Standard value for reflective indices of water

= 1.4655  5.5964  10 5 = 1.50-1.54 = 1.3721 1.2356  10 4 = 1.333

Discussion: This was the first practical on this course.It was a New experience work in Optical Labaratory. It was difficult to adjust the Spectrometer by using Schuster’s method.But finally adjusted it well. Before Use Prism we should clean it.If not we couldn’t get some observations clearly. In first part I think I did some mistake on measure T1 & T2. So I think that’s the reason for show some difference between my practical values & standard values. Demarcation Line is separate bright & dark areas.It was very hard to find Demarcation line on part (II).But finally I found it.