Question

In a double-slit experiment, the slits are illuminated by a monochromatic, coherent light source having a wavelength of 609 nm. An interference pattern is observed on the screen. The distance between the screen and the double-slit is 1.69 m and the distance between the two slits is 0.118 mm. A light wave propogates from each slit to the screen. What is the path length difference between the distance traveled by the waves for the sixth-order maximum (bright fringe) on the screen?

Answer 1

Answer:

$\Delta x = 3.65 \mu m$

Explanation:

As we know that the sixth order maximum will have path difference given as

$\Delta x = N\lambda$

here we know that

N = order of maximum

$\lambda = 609 nm$

now we have

$N = 6$

so we know that

$\Delta x = 6(609 nm)$

$\Delta x = 3.65 \mu m$