Question

A mixture of polarised and unpolarised light is incident on a polariser. As the polariser is

rotated through 360°, it is found that the minimum and the maximum of the transmitted

intensity is in the ratio 1 : q. The ratio of intensities of polarised to unpolarised light in

the incident beam is

(A) 1

2

(q − 1) : 1 (B) 1 : 1

2

(q − 1) (C) 1

2

q : 1 (D) 1 : 1

2

q​

Answer 1

Answer: (A) 1/2(q-1) : 1 .

Step-by-step explanation:

1) Let the intensity of polarized and unpolarized light be I_1 and I_2 respectively.

When polarized rotated by 360° , it comes across many different types of transmitted intensities .

Transmitted intensity of unpolarized beam of light will always be

$\frac{I_2}{2}$

no matter what the angle of polarized is!

2) But, for the case of polarized light, it's intensity changes.

When direction of polarized is parallel to that of polarized light, then transmitted intensity will be I_1.

But, when direction of polarizer is perpendicular to polarized light, then transmitted intensity will be 0

3) According to above ideas,

$\frac{I_{min}}{I_{max}}=\frac{1}{q} \\ \\=> \frac{I_2/2}{I_2/2+I_1}=\frac{1}{q}\\ \\=>q=1+\frac{2I_1}{I_2}\\ \\=>q-1=\frac{I_1}{2I_2} \\ \\=>\frac{I_1}{I_2}=\frac{(q-1)/2}{1}$

Hence, Above is our required ratio.