Question

In a double-slit experiment, the slits are illuminated by a monochromatic, coherent light source having a wavelength of 507 nm. An interference pattern is observed on the screen. The distance between the screen and the double-slit is 1.32 m and the distance between the two slits is 0.112 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 fifth-order maximum (bright fringe) on the screen

Answer 1

Answer:

The path difference is $2.53 \times 10^{-6}$ m

Explanation:

Given:

Wavelength of light $\lambda = 507 \times 10^{-9}$ m

Distance between slit and screen $D = 1.32$ m

Distance between two slit $d = 0.112 \times 10^{-3}$ m

Order of interference $n = 5$

From the formula of interference of light,

   $d\sin \theta = n\lambda$

Where $d \sin \theta =$ path difference, $n =$ order of interference

Here we have to find path difference,

 Path difference $= 5 \times 507 \times 10^{-9}$ m

 Path difference $= 2.53 \times 10^{-6}$ m

Therefore, the path difference is $2.53 \times 10^{-6}$ m