Question

A polaroid examines two adjacent plane
polarised beams A and B whose planes of
polarisation are mutually perpendicular. In the
first position of the analyser, beam B shows
zero intensity. From this position a rotation 30°
shows that the two beams have same intensity.
The ratio of intensity of the two beams IA and
IB
1) 1:3
2) 3:1
3) sqrt3:1
4) 1:sqrt3​

Answer 1

Solution :

Let θ be the angular position of the beam A when intensity of beam B = 0.

Let intensity of beam:

$\implies$ $\sf{A = I_{0A}}$

$\implies$ $\sf{B = I_{0B}}$

Appeared intensity of:

$\implies$ A for a particular angle = $\sf{I_{A}}$

$\implies$ B for a particular angle = $\sf{I_{B}}$

Since:

$\implies$ $\huge{\boxed{\sf{I_{B} = 0}}}$

By law of Malus,

$\implies$ $\sf{I_{B} = I_{OB} cos^{2} (\theta +90) = 0 }$

$\implies$ $\sf{(-sin (\theta ))^{2} = 0}$

$\implies$ $\sf{\theta = 0}$

Thus, original position of the beam A was at 0° of the polaroid. Now, after a 30° of rotation of the polaroid $\sf{I_{A} = I_{B}}$.

That is:

$\implies$ $\sf{I_{A} = I_{OA} cos^{2} (\theta +30\degree)}$

$\implies$ $\sf{I_{A} = I_{OB} cos^{2}(\theta+30\degree+90\degree )}$

$\implies$ $\sf{I_{A} = I_{B}}$

Since, θ = 0

$\implies$ $\sf{I_{OA} cos^{2} (30\degree) = I_{OB} cos^{2} (120\degree)}$

$\implies$ $\sf{I_{OA} \times \frac{3}{4} = I_{OB} \times \frac{1}{4}}$

$\implies$ $\sf{\frac{I_{OA}}{I_{OB}} = \frac{1}{3}}$

Ratio of intensity of beams A and B are 1:3.

Therefore,

Correct option: (1) 1:3

________________

Answered by: Niki Swar, Goa❤️

Answer 2

Ratio of intensity of beams A and B are 1:3.

Therefore,

Correct option: (1) 1:3