Question
In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength, If wavelength of red light is $$6600\,\mathop {\text{A}}\limits^ \circ ,$$ then wavelength of first maximum will be :
A.
$$3300\,\mathop {\text{A}}\limits^ \circ $$
B.
$$4400\,\mathop {\text{A}}\limits^ \circ $$
C.
$$5500\,\mathop {\text{A}}\limits^ \circ $$
D.
$$6600\,\mathop {\text{A}}\limits^ \circ $$
Answer :
$$4400\,\mathop {\text{A}}\limits^ \circ $$
Solution :
In a single slit experiment,
For diffraction maxima, $$a\sin \theta = \left( {2n + 1} \right)\frac{\lambda }{2}$$
and for diffraction minima, $$a\sin \theta = n\lambda $$
According to question,
$$\eqalign{
& \left( {2 \times 1 + 1} \right)\frac{\lambda }{2} = 1 \times 6600\,\,\left( {\because {\lambda _R} = 6600\,\mathop {\text{A}}\limits^ \circ } \right) \cr
& \lambda = \frac{{6600 \times 2}}{3} = 4400\,\mathop {\text{A}}\limits^ \circ \cr} $$