Physics 1 – Spherical Refraction
Aims
 To investigate refraction on a semi-circular plastic block.

To calculate the refractive index of the plastic block.

To setup experimental apparatus correctly and obtain suitable measurements verifying predictions.
Part 1: Tutorial Question (20 mins)
Case A A semi-circular plastic block with radius 37 mm and a refractive index of 1.5 is placed in the path of a laser beam as shown below in Figure
1.1.
1.1 Determine the image distance s’ at which the rays focus from the semi-circular block. s

Laser rays n
A n
B n
A
Vertex fo r
Case B sp
The plastic block is now positioned as shown in Figure 1.2. he
1.2
Why will the new image distance, s
2
’ ri
, be different from Figure 1.1?
Figure 1.1
1.3
Calculate the distances of the imaginary image s
1
’ ca l s
2
’
using the same parameters as above. re fr s
1
’ ac s
2
’ ti ng su rf ac
Laser rays n
A n
B n
A
Vertex 2 Vertex 1 fo r sp Figure 1.2 he
Hints: ri ca st boundary, then move on to what happens at the 2 nd re fr
1 ac ti ng su

Physics 1 – Spherical Refraction
Remember the relationship; n s
A  n s
B

 n
B

R n
A for light refracted at a spherical boundary of radius R between medium A (refractive index n
A
) and medium B (refractive index n
B
).
From the lectures/textbook you will need to recall the necessary sign convention!
Part 2: Taking Measurements (25 mins)

Setup the apparatus for Case A . Ensure the switch on the RayZer Lightbox is on the 3 ray position, and use graph paper to draw the points of incidence on the block, departure from the block and where they focus.
2.1
Next, remove the plastic block and join up the lines to produce a schematic similar to Figure 1.1. Draw in the vertex with a dashed line and let it be the line x = 0. This will help you to put the correct signs in front of your units.

Measure the image distance.
2.2
Use this image distance to determine the refractive index of the plastic block and discuss how your results compare with prediction.

Now, setup the apparatus for Case B and use graph paper to draw the points of incidence, departure, and where they focus.

Draw in the vertices to produce a schematic similar to Figure 1.2.
2.3
Using the same value of refractive index you obtained previously, predict the imaginary image distance from Vertex 1.

Compare this value with the distance you found.
 Mark the point of the imaginary object and determine its distance.

Now, using the object distance, predict the distance of the image formation and compare this with the distance you measured.
Hints:
Be careful to put the correct signs for the values of s and s’
.
Part 3: Brainstorm (5 mins)
3.1(a) Write down any reasons why your results show deviation from that of the prediction.
(b) What would happen, in both cases, if the plastic block was now replaced by another with a larger refractive index?
(c) What would happen to the image distances if the red laser light was changed to green?
Hints:
Are the laser rays completely parallel?
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Physics 1 – Spherical Refraction
What would happen if the separation of the laser rays was reduced?
Further Work
The following questions are related to the topic covered by this experimental tutorial.

Exercise book questions L26 – L28.

Mastering Physics: Optics 1 (last part)

Take a look at http://en.wikipedia.org/wiki/Negative_refractive_index for some recent developments of exotic materials with negative refractive index!
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Physics 1 – Spherical Refraction
Demonstrators' Answers, Hints, Marking Scheme and Equipment List
Marking Scheme
Section
1.1
1.2
1.3
Mark
1
1
2
2.1
2.2
2.3
3.1
Discretionary mark
TOTAL
1
1
1
1
2
10
Answers
1.1
1 .

5

1 s '


0 .
5
37
 s '


37
0 .
5
 
74 mm , i.e. 74 mm to the left of the curved side of the block.
1.2
The rays are being refracted twice. When the rays meet the 1 st
boundary they are refracted, then when they meet the 2 nd
boundary they are converging from an imaginary object, no longer parallel rays from infinity.
1.3
n s a
1
 n s
1 b
'
 n b
 n a
R


1
 s
1
'

1 .
5 s
1
'



0

111 mm
.
5
37 s
2

111

37

74 mm n a s
2
 n b s
2
'

1 .
5
74

1
 n b
 n a s
2
'

R

0 .
5

 s
2
'

49 .
3 mm
Part 2 see example on graph paper
3.1(a) The rays are not perfectly parallel because they diverge ~ 1mm in 100mm.
The separations of the rays on either side of the central ray are not exactly the same.
(b) The refraction angles would increase causing a shorter image distance in both cases.
4

Physics 1 – Spherical Refraction
(c) Green light has a shorter wavelength so is refracted more so the image distance would decrease.
5

Physics 1 – Spherical Refraction
6

Physics 1 – Spherical Refraction
Equipment list:
Laser-board with fitted laser and power supply
Box of Perspex blocks
Graph paper
Protractor
Ruler
7