Answer to Question #171612 in Optics for Mohammed Musa Dabo

Question

In a young double slit experiment the separation is 0.05cm and the distance between the double slit and screen is 200cm when blue light is used , the distance of the interference pattern is 0.13cm

Answer 1

Given,

Seperation d=0.05 cm

D=200 cm

x=0.13 ,

n=1

For bright fringes, $x=\dfrac{nλD}{d} or λ=\dfrac{xd}{nD}$

Substituting the values, we get

$\lambda=\dfrac{0.13\times 0.05}{ 1\times 200}=\dfrac{65\times 10^{-4}}{200}=0.325\times 10^{-4}cm$