Question

In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \(P_{1}\) and \(P_{2}\) are orthogonal to each other. The polarizer \(P_{3}\) covers both the slits with its transmission axis at \( 45^{\circ}\) to those of \(\mathrm{P}_{1}\) and \(P_{2}\). An unpolarized light of wavelength \(\lambda\) and intensity \(\mathrm{I}_{0}\) is incident on \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\). The intensity at a point after \(P_{3}\) where the path difference between the light waves from \(\mathrm{s}_{1}\) and \(\mathrm{s}_{2}\) is \(\frac{\lambda}{3}\), is

 

(1) \(\frac{I_{0}}{2}\)

(2) \(\frac{I_{0}}{4}\)

(3) \(I_{0}\)

(4) \(\frac{I_{0}}{3}\) 

Answer 1

Correct option is : (2) \(\frac{I_{0}}{4}\)  

Incident amplitude is \(A_0\) then

\(I_0 \propto A_0^2\) 

 

Now, path difference is \(\frac{\lambda }{3}\) so phase difference is \(\frac{2\pi }{2}\) 

\(A^2 = A_1^2 + A_2^2 + 2A_1A_2\ cos \theta \) 

\( = \frac{A_0^2}{4} + \frac{A_0^2}{4} + 2\frac{A_0^2}{4}.\cos (\frac{2 \pi}{3})\) 

\(\Rightarrow A^2 = \frac{A_0^2}{4}\) 

so, \(I = \frac{I_0}{4}\)