Question

in a young's double slit experiment the intensity of the resultant wave at point P on the screen is I where the path difference between the waves from coherent sources S1 and S2 is λ. then find the intensity of the resultant wave at a point where the path difference is λ/4

Answer 1

Phase difference,
$\phi =\frac{2\pi }{\lambda }\Delta$
Intensity,
$I=4I_0{\cos }^2\frac{\phi }{2}$
Therefore,
$$\begin{gathered} \frac{I_1}{I_2}=\frac{{\cos }^2{\displaystyle \frac{{\phi }_1}{2}}}{{\cos }^2\frac{{\phi }_2}{2}}=\frac{{\cos }^2\left({\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2\pi }{\lambda }}\times \lambda \right)}{{\cos }^2\left(\frac{1}{2}\times \frac{2\pi }{\lambda }\times {\displaystyle \frac{\lambda }{4}}\right)}=\frac{{\cos }^2\pi }{{\cos }^2{\displaystyle \frac{\pi }{4}}}=2 \\ \Rightarrow \frac{I}{I_2}=2 \\ \Rightarrow I_2=\frac{I}{2} \end{gathered}$$