For a grating that contains 1400 blazes/mm, if angle of incidence = 48 ∘ and n = 2, then wavelength that will appear at angle of reflection 20 ∘ is:A. 0.748 nmB. 374 nmC. 187 nmD. 748 nm

Question

For a grating that contains 1400 blazes/mm, if angle of incidence   =      48    ∘   and   n  =  2, then wavelength that will appear at angle of reflection       20    ∘   is:A. 0.748 nmB. 374 nmC. 187 nmD. 748 nm

toliwask · Accepted Answer

This is a problem related to the diffraction of the light through a diffraction grating.The formula for the reflection grating can be written as,  n  λ  =  d  (  sin  ⁡      θ    i    +  sin  ⁡      θ    d    )Where,n - Order of diffraction  λ - Wavelength of light      θ    i   - Angle of incidence      θ    d   - Angle of diffractiond − spacing between each linData:  n  =  2    λ  =  ?        θ    i    =      48    ∘          θ    d    =      20    ∘      d  =      1          1400       blazes/mm          =  0.714  ×      10          −      3         mm    =  0.714  ×      10          −      6         mFormula:  n  λ  =  d  (  sin  ⁡      θ    i    +  sin  ⁡      θ    d    )Solution:Using the given formula and substituting the values, we get,  n  λ  =  d  (  sin  ⁡      θ    i    +  sin  ⁡      θ    d    )    2  ×  λ  =  0.714  ×      10          −      6        ×  (  sin  ⁡      48    ∘    +  sin  ⁡      20    ∘    )    2  ×  λ  =  0.714  ×      10          −      6        ×  (  0.743  +  0.342  )    2  ×  λ  =  0.714  ×      10          −      6        ×  1.085    λ  =            0.714      ×              10                  −          6                    ×      1.085        2      λ  =  3.873  ×      10          −      7         m    λ  =  387.3  ×      10          −      9         m    λ  =  387.3   nmAnswer:Only the option B matches our result.

For a grating that contains 1400 blazes/mm, if angle of incidence =48^circ and n=2

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