Question #162536

Light of wavelength 587.5 nm illuminates a slit of width 0.75 mm.

(a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.85mm from the central maximum?

(b) Calculate the width of the central maximum.


Expert's answer

Answer

a)

λ=587.5×10−9m\lambda=587.5\times10^{-9}m

a=0.75×10−9m0.75\times10^{-9}m

y1=0.85×10−3my_1=0.85\times10^{-3}m

distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.85mm from the central maximum

L=y1aλL=y_1\frac{a}{\lambda}

=0.85×10−3×0.75×10−9587.5×10−9\frac{0.85\times10^{-3}\times0.75\times10^{-9}}{587.5\times10^{-9}}

=1.1m


b) the width of the central maximum.

X=2DλdX=\frac{2D\lambda}{d}

=2×1.1×587.5×10−90.85×10−3=1.5mm=\frac{2\times1.1\times587.5\times10^{-9}}{0.85\times10^{-3}}\\=1.5mm


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