Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

A double-slit experiment is performed with light of wavelength 550 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?

Sagot :

okpalawalter8

Answer:

[tex]d_2=1.5*10^-3m[/tex]

Explanation:

From the question we are told that:

Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]

Space 1 [tex]d_1=2.3*10^{-3}[/tex]

Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]

Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by

 [tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]

 [tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]

 [tex]d_2=1.5*10^-3m[/tex]

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.