A plane diffraction grating is illuminated by a source that emits two spectral lines of wavelengths 420nm and 600nm. Show that the 3rd order of the line of one of these wavelengths is diffracted through a greater a greater angle than the 4th order of the other wavelength.
Answer
"\\lambda_1=420nm\\\\\\lambda_2=600nm"
n1=3
n2=4
Using bragg equation
"2dsin\\theta=n\\lambda"
Here interplaner distance and wavelength is different.
So angle relationship
"\\frac{sin\\theta_1}{sin\\theta_2 }=\\frac{n_1\\lambda_1}{n_2 \\lambda_2 }=\\frac{3\\times420}{4\\times600}"
"{sin\\theta_1}=0.525\\times sin\\theta_2"
Which means "\\theta_1" Is lesser.
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