Question

In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright fringe on a flat screen is 0.0176 m, when the light has a wavelength of 425 nm. Assume that the angles are small enough so that is approximately equal to . Find the separation y when the light has a wavelength of 614 nm.

Answer 1

Answer:

Y = 0.0254 m = 25.4 mm

Explanation:

The formula for the fringe spacing in Young's Double-slit experiment is given by the following formula:

$Y = \frac{\lambda L}{d}$

where,

Y = fringe spacing = 0.0176 m

λ = wavelength = 425 nm = 4.25 x 10⁻⁷ m

L = Distance between screen and slits

d = slit separation

Therefore,

$\frac{0.0176\ m}{4.25\ x\ 10^{-7}\ m} = \frac{L}{d}\\\\\frac{L}{d} = 41411.76$

Now, for:

λ = 614 nm = 6.14 x 10⁻⁷ m

$Y = (6.14\ x\ 10^{-7}\ m)(41411.76)$

Y = 0.0254 m = 25.4 mm