Answer to Question #113466 in Optics for Hayley

Question

Geraldine uses a 439-nm coherent light source and a double slit with a slit separation of 26.0 μm to display three interference maxima on a screen that is 20.0 cm wide. If she wants to spread the three bright fringes across the full width of the screen, from a minimum on one side to a minimum on the other side, how far from the screen should she place the double slit?

Answer 1

As per the given question,

wavelength of the wave $(\lambda)=439nm$

separation between the slit $(d)=26\mu m$

width of the interference maxima =20.0 cm

Hence $20/2=10cm$

total number of interference maxima (m)=3

Let the initial distance between the slit and the screen is D

now, we know that $y=\dfrac{m\lambda D}{d}$

$D=\dfrac{yd}{m\lambda}=\dfrac{10\times 10^{-2}\times 26\times 10^{-6}}{3\times 439\times10^{-9}}=1.97m \simeq 2m$

Now, for the second case y=20,

m=3

$(d)=26\mu m$

$D=\dfrac{yd}{m\lambda}=\dfrac{20\times 10^{-2}\times 26\times 10^{-6}}{3\times 439\times10^{-9}}=3.95m \simeq 4m$