**1 Answer**

written 7.0 years ago by | modified 5.5 years ago by |

Then continue as follows:-

For $m^{th}$ principal maximum,

$(a + b) sin θ = mλ → sin θ = \frac{mλ}{a + b}$

For minimum intensity in single slit

$a sin θ = nλ → sin θ = \frac{nλ}{a}$

If the two conditions are satisfied simultaneously,

$\frac{mλ}{a + b} = \frac{nλ}{a}$

$m = (\frac{a + b}{a}) n $

...This is the required expression for missing spectra.

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Sometimes it happen that the first order spectrum is visible , second order is not visible ,third order is again visible and so on i,e some spectra’s are missing in the grating

We know , The resultant intensity is given by

$I_θ$ =[ $I_m$ $sin^2$ α ) / $α^2$ ][ $sin^2$ Nβ) / $sin^2$ β)]

$I_θ$ = $I_s$ X $I_g$

If $I_s$ =o then $I_θ$ =o irrespective of what value $I_g$ have.

i,e the condition of minima for single slit and for maxima in grating are satisfied together at a point on the screen. Then a particular maximum of order ‘m’ will be missing in the grating

for grating maxima is (a+b) sin θ = mλ ……………..(1)

for minima in single slit asinθ = nλ …………………(2)

(a+b)sinθ / asinθ= mλ / nλ

(a+b) / a= m / n

m= [(a+b) / a] n condition for absent spectra

Consider a case where b=a .i.e width of opaque part is same as that of slit.

Then m= [ (a+a)/a] n

Therefore, m =2n.

That means 2,4,6,8……..these orders will be missing in the spectra.