What Will The Fringe Spacing Be If The Light Is Changed To A Wavelength Of 400 Nm

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The Interference of Light:

• Thread starter frederickcan
• Offset date Jul 28, 2009

Homework Argument

A double-slit experiment is performed with light of wavelength 600 nm. The bright interference fringes are spaced ane.8 mm autonomously on the viewing screen. What will the fringe spacing exist if the light is changed to a wavelength of 400 nm?

Homework Equations

Fringe spacing equation
delta y = wavelength ten L / d

The Endeavour at a Solution

I tried to solve the fringe spacing for the wavelength of 600 nm by:

600 ten 10^-9 thousand / .0018 yard = 3.33E-4

and then

plugged in the 3.33E-4 to solve what the spacing would be for 400 nm by:

dividing 400 x 10^-nine yard / three.33E-4 = 1200000

I recall I'm interpreting the fringe spacing equation wrong, and I'm not sure where.

Answers and Replies

Check your math; your equations are right, only you fabricated a footling calculation mistake.

The but thing is I'k dislocated about L. The question didn't provide a value for Fifty, and I have ii unknowns (L, and the fringe spacing) then I feel similar I'm using the wrong equation. However, because I couldn't find a better equation I set L to one.

(Also, thank you for your help.)

In the problem L and d remain the same. So you tin write equation for 2 wavelengths.
λ1 = x1*L/d....(one)
λ2 = x2*L/d....(2) From the start equation detect L/d.
Substitute this value in the 2nd equation.

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