A) red
B) blue
C) violet
D) green
Correct Answer: C
Solution :
When a wave of wavelength\[\lambda \]falls at an angle of incidence i on a film of refractive index\[\mu \]and thickness t, then the condition for constructive interference in the reflected system is \[2\mu t\cos r=\left( m+\frac{1}{2} \right)\lambda ,m=0,1,\,2,\,3\]where, r is the angle of refraction in the film. For normal incidence\[i=0,\] hence,\[~r=0.\] Also\[\mu =1\] for air \[\therefore \]\[2t=\left( m+\frac{1}{2} \right)\lambda =\frac{\lambda }{2},\frac{3\lambda }{2},\frac{5\lambda }{2},...\]etc or \[\lambda =\frac{2t}{\left( m+\frac{1}{2} \right)}=4t,\frac{4t}{3},\frac{4t}{5},...\]etc Now Therefore, \[\lambda =11600\,\overset{\text{o}}{\mathop{\text{A}}}\,,3867\overset{\text{o}}{\mathop{\text{A}}}\,,2320\overset{\text{o}}{\mathop{\text{A}}}\,,....,\]etc. Wave of \[1=11600\,\overset{\text{o}}{\mathop{\text{A}}}\,\]is in the infrared r \[t=0.00029\text{ }mm=2.9\times {{10}^{-5}}cm=2900\overset{\text{o}}{\mathop{\text{A}}}\,\] egion and wave of \[\lambda =11600\overset{\text{o}}{\mathop{\text{A}}}\,\]is in the ultraviolet. These waves are not visible. Hence, the visible wave in the reflected system has a wavelength\[\lambda =2320\,\overset{\text{o}}{\mathop{\text{A}}}\,,\] which is close to violet light\[(\lambda =3800\overset{\text{o}}{\mathop{\text{A}}}\,).\] Hence, the correction option is [b].
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