A) 2 mm
B) 2 cm
C) 2 m
D) 2 km
Correct Answer: A
Solution :
| [a] Let d in cm be the thickness of air column = thickness of vacuum column (given). Number of waves of wavelength \[\lambda =6000\overset{\text{o}}{\mathop{\text{A}}}\,=6\times {{10}^{-5}}cm\] in a thickness d cm in vacuum is \[{{n}_{v}}=\frac{d}{\lambda }\] |
| Since the refractive index of air \[\mu =1.0003,\] the wavelength in air will be \[{{\lambda }_{a}}=\frac{\lambda }{\mu }\] |
| Therefore, number of waves of wavelength \[{{\lambda }_{a}}\] of air is \[{{n}_{a}}=\frac{d}{{{\lambda }_{a}}}=\frac{d\mu }{\lambda }\] |
| Given that \[{{n}_{a}}+1={{n}_{v}}\] |
| Hence \[\frac{d\mu }{\lambda }+1=\frac{d}{\lambda }\] |
| \[d=\frac{\lambda }{\mu -1}\] |
| \[=\frac{6\times {{10}^{5}}cm}{1.0003-1}\] |
| \[=0.2cm=2mm\] |
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