A) \[0.16\,\times {{10}^{-6}}\,m,\,\,6.25\,\times {{10}^{14}}\,Hz\]
B) \[0.36\,\times {{10}^{-6}}\,m,\,\,6.25\,\times {{10}^{14}}\,Hz\]
C) \[0.36\,\times {{10}^{-6}}\,m,\,\,3.25\,\times {{10}^{14}}\,Hz\]
D) \[0.26\,\times {{10}^{-6}}\,m,\,\,3.25\,\times {{10}^{14}}\,Hz\]
Correct Answer: B
Solution :
\[\,{{\lambda }_{m}}=\frac{{{\lambda }_{a}}}{\mu }=\frac{0.48\times {{10}^{-6}}}{4/3}m=0.36\times {{10}^{-6}}m\] Frequency of light in air \[v=\frac{c}{{{\lambda }_{a}}}=\frac{3\times {{10}^{8}}}{0.48\times {{10}^{-6}}}=6.25\times {{10}^{14}}Hz\] On passing through water (medium), frequency does not change. Hence, it will be \[6.25\times {{10}^{14}}\,Hz\]in water.
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