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You illuminate a slit with a width of 0.0549 mm with a light of wavelength 735 nm and observe the resulting diffraction pattern on a screen that is situated 2.85 m from the slit. What is the width, in centimeters, of the pattern's central maximum?

Respuesta :

Answer:

Explanation:

Given

slit width [tex]d=0.0549\ mm[/tex]

wavelength of light [tex]\lambda =735\ nm[/tex]

Screen is placed at a distance of [tex]D=2.85\ m[/tex]

The angle for the first minima is given by

[tex]\sin \theta _1=\frac{m\lambda }{d}[/tex]

for m=1

[tex]\sin \theta _1=\frac{1\times 735\times 10^{-9}}{0.0549\times 10^{-3}}[/tex]

[tex]\sin \theta _1=0.0133[/tex]

and one half of central maxima is given by

[tex]Y=D\sin \theta _1[/tex]

[tex]Y=2.85\times 0.0133[/tex]

[tex]Y=0.0381\ m[/tex]

Width of central maxima [tex]=2Y[/tex]

[tex]Width=0.0762\ m\approx 7.62\ cm[/tex]                                

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