• # question_answer A mixture of violet light of wavelength $3800\,\overset{\text{o}}{\mathop{\text{A}}}\,$ and blue light of wavelength $4000\,\overset{\text{o}}{\mathop{\text{A}}}\,$ is incident normally on an air film of 0.00029 mm thickness. The colour of the reflected light is A)  red                B)  blueC)  violet                          D)  green

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].