In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of P1 and P2 are orthogonal to each other. The polarizer P3 covers both the slits with its transmission axis at 45 to those of P1 and P2. An unpolarized light of wavelength and intensity I0 is incident on P1 and P2. The intensity at a point after P3 where the path difference between the light waves from s1 and s2 is 3, is [figure]
- I0
- I03
- I04
- I02
Answer
(3) I04
Solution
P1,P2 each pass I02; P3 (at 45 ) passes I1=I2= I02 245 = I04 — now the waves interfere.
= 2 3=120
[figure]
I=I1+I2+2 I1I2 120 = I04+ I04- I04
= I04
Answer: (3)