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

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Answer:

The path difference is [tex]2.53 \times 10^{-6}[/tex] m

Explanation:

Given:

Wavelength of light [tex]\lambda = 507 \times 10^{-9}[/tex] m

Distance between slit and screen [tex]D = 1.32[/tex] m

Distance between two slit [tex]d = 0.112 \times 10^{-3}[/tex] m

Order of interference [tex]n = 5[/tex]

From the formula of interference of light,

   [tex]d\sin \theta = n\lambda[/tex]

Where [tex]d \sin \theta =[/tex] path difference, [tex]n =[/tex] order of interference

Here we have to find path difference,

 Path difference [tex]= 5 \times 507 \times 10^{-9}[/tex] m

 Path difference [tex]= 2.53 \times 10^{-6}[/tex] m

Therefore, the path difference is [tex]2.53 \times 10^{-6}[/tex] m