In the phenomenon of double refraction using calcite crystal, it has been found that 0-ray and E-ray are plane polarised, because when a tourmaline crystal is placed in the path of 0-ray and E-ray, then on rotating the crystal, intensity of both the images changes. If intensity of one image (extraordinary) decreases then intensity of other image (ordinary) increases. In complete rotation, both the images alternatively can be extinguished at two places. If clearly shows that both 0-ray and E-ray are plane polarised with their planes perpendicular to each other.
Huygen’s in 1678, demonstrated polarisation of light by double refraction. In his experiment, he passed a beam of light through a pair of calcite crystals and made the following observations:
(i) When principal sections of both the crystals are parallel i.e., e = 0 as shown in Figure 6.9a. Two images 01 and E1 are formed due to a ray falling normally on the surface AB of first crystal. This. is because, a ray falling normally on first crystal gets splitted into 0-ray and £-ray. The 0-ray transverse second crystal without any deviation whereas E-ray transverse second crystal along the path parallel to that in the first crystal. Thus two rays represented by 01 and E1 emerge exactly parallel with the separation twice as in case of first crystal. ·
(ii) If the second crystal is rotated, each of the two rays 0-ray and E-ray suffers double refraction in the second crystal and four images are observed. The old images 01 and E1 become dimmer and in between them two new faint images 02 and E2 are formed. On rotation of second crystal, the images 01 and 02 remains stationary while E1 and £2 rotate in a circular path around 01 and 02 respectively. In addition to it, intensity of old image 01 and £1 decreases whereas that of new images 02 and E2 increases. When principal section of second crystal makes an angle of 45° with the principal section of first crystal, all the images are of equal intensity as shown in Figure 6.9(b).
(iii) When e = 90°, the old images 01 and E1 disappear and new images 02 and E2 acquires
maximum intensity as shown in Figure 6.9(c).
(iv) On further rotation, when e = 135°, again four images of equal intensity are observed as shown in Figure 6.9(d).
(v) Ate= 180°, the principal section of two planes are parallel but their optic axis are oriented in opposite direction as shown in Figure 6.9(e), so the images 02 and E2 disappear and images 01 and E1 will superimpose to form a single.
(vi). If rotation continues from180o to 360°, all the above changes takes place in reverse direction.
Thus, it has been derived experimentally that first crystal provides polarised lights as 0-ray and E-ray.
Explanation: The above observations can be explained physically as:
Let e be the angle between principal section CD and C’D’ of two crystals at any instant.
When an ordinary ray enters first crystals, it splits up into two plane polarised component as 0-ray and E-ray. The vibrations of 0-ray are perpendicular to principal section CD. In Figure 0-ray and E-ray are represented by PO and PE respectively and each having same amplitude say a. Both 0-ray and £-ray enter second crystal and gets splitted further into two component each due to double refraction. The components of x-ray are: ordinary component E2 of amplitude a cos θ along PO1 perpendicular to principal section C’D’ of crystal-2 and extraordinary component £ of amplitude a sin e along P£1 along the principal section C’D’ of crystal-2. θ
As Intensity = (Amplitude)2
So Intensity of each of O1 and E1 = a2 cos2 θ
and Intensity of each of O2 and E2 =a2 sin2 θ
When e θ= 0 or θ = 180°
a2 cos2 e = a2 i.e., O1 andE1 has maximum intensity.
And a2 sin2 θ =0 i.e., O2 and E2 disappear.
When e = 45° or e = 135°
A2 cos 2 θ =a2/2
A2 sin 2 θ =a2 /2 i.e; O1E1 and O2 E2 all have same intensity
When θ =900
A2 sin2 θ =a2 i.e; O1 E1 disappear
And a2 sin 2 θ =a2 i.e; O1 E2has maximum intensity
Thus all the observations made by Huygen can be explained physically and for all positions the sum of intensities of two components is a2 cos2 e + a2 sin2 e = a2, which is equal to intensity of incident beam.