Question

The intensity on the screen at a certain point in a double-slit interference pattern is 64.0% of the maximum value. (a) What minimum phase difference (in radians) between sources produces this result? (b) Express this phase difference as a path difference for 486.1-nm light.

Answer 1

(a) we know that

$I=I_{max}co{s}^2\left(\frac{\pi dsin\theta }{\lambda }\right)$

with $\varphi =\frac{2\pi }{\lambda }dsin\theta .$ This gives

$\frac{I}{I_{max}}=co{s}^2\left(\frac{\varphi }{2}\right)$

Therefore,

$\varphi =2co{s}^{-1}\sqrt{\frac{I}{I_{max}}}=2co{s}^{-1}\sqrt{0.640}=1.29rad$