Question #164081

  A source emitting light of two wavelengths is viewed through a grating spectrometer set at normal incidence. When the telescope is set at an angle of 20° to the incident direction, the second-order maximum for one wavelength is seen superposed on the third-order maximum for the other wavelength. The shorter wavelength is 400nm. Calculate the longer wavelength and the number of lines per centimeter in the grating. At what other angles, if any, can the superposition of two orders be seen using this source?



1
Expert's answer
2021-02-19T10:31:52-0500

Given quantities:

the longer vawelength is belong to light which have a bigger nth-order maximum

θ=20o\theta = 20^o m1=2m_1 = 2 λ2=400nm\lambda_2= 400nm m2=3m_2 = 3 λ1?\lambda_1 - ?

m1λ1=dsinθm_1\lambda_1= dsin\theta m1λ1=m2λ2λ1=m2λ2m1m_1\lambda_1= m_2\lambda_2 \to \lambda_1 = \large\frac{m_2\lambda_2}{m_1} =600nm= 600nm

m2λ2=dsinθm_2\lambda_2= dsin\theta

the number of lines per centimeter in the grating l=1cml = 1cm

N=ld=lsinθm1λ1=102sin20o2600109N = \large\frac{l}{d} = \frac{lsin\theta}{m_1\lambda_1} = \frac{10^{-2}*sin20^o}{2*600*10^{-9}} =2850= 2850


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