A) \[4400\,\,{AA}\]
B) \[6600\,\,{AA}\]
C) \[2000\,\,{AA}\]
D) \[3500\,\,{AA}\]
Correct Answer: A
Solution :
In a single slit diffraction experiment, position of minima is given by \[d\sin \theta =n\lambda \] So for first minima of red \[\sin \theta =1\times \left( \frac{{{\lambda }_{R}}}{d} \right)\] and as first maxima is midway between first and second minima, for wavelength \[{\lambda }'\], its position will be \[d\sin {\theta }'=\frac{{\lambda }'+2{\lambda }'}{2}\Rightarrow \sin {\theta }'=\frac{3{\lambda }'}{2d}\] According to given condition \[\sin \theta =\sin {\theta }'\] \[\Rightarrow {\lambda }'=\frac{2}{3}{{\lambda }_{R}}\] so \[{\lambda }'=\frac{2}{3}\times 6600=440\,nm\]\[=4400\,{AA}\]
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