A beam of ordinary unpolarised light passes through a tourmaline crystal C1 and then it passes through another tourmaline crystal C2, which is oriented such that its principal plane is parallel to that of C2. The intensity of emergent light is I0. Now C2 is rotated by 60° about the ray. The emergent ray will have an intensity

Question

A beam of ordinary unpolarised light passes through a tourmaline crystal C1 and then it passes through another tourmaline crystal C2, which is oriented such that its principal plane is parallel to that of C2. The intensity of emergent light is I0. Now C2 is rotated by 60° about the ray. The emergent ray will have an intensity (A) 2 I0 (B) I0 / √2 (C) I0 / 4 (D) I0 / 2

Tardigrade Admin · Accepted Answer

Intensity of light from C2 =I0 On rotating the crystal through 60°, I=I0 cos 2 60° (Malus law) =I0((1/2))2=I0 / 4

$2 I_{0}$

$I_{0} / 2$

$I_{0} /4$

$I_{0} / 2$

Intensity of light from $C_{2} = I_{0}$
On rotating the crystal through $6 0^{\circ}$ ,
$I = I_{0} cos^{2} 6 0^{\circ}$ (Malus law)
$= I_{0} (\frac{1}{2})^{2} = I_{0} /4$

2I0​

I0​/2​

I0​/4

I0​/2​

Intensity of light from C2​=I0​ On rotating the crystal through 60∘, I=I0​cos260∘ (Malus law) =I0​(21​)2=I0​/4

$2 I_{0}$

$I_{0} / 2$

$I_{0} /4$

$I_{0} / 2$

Intensity of light from $C_{2} = I_{0}$
On rotating the crystal through $6 0^{\circ}$ ,
$I = I_{0} cos^{2} 6 0^{\circ}$ (Malus law)
$= I_{0} (\frac{1}{2})^{2} = I_{0} /4$