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  • The difference of speed of light in the two media and <img alt="\mathrm{B}\left(v_{\mathrm{A}}-v_{\mathrm{B}}\right)" src="ht

The difference of speed of light in the two media A and \mathrm{B}\left(v_{\mathrm{A}}-v_{\mathrm{B}}\right) is 2.6 \times 10^{7} \mathrm{~m} / \mathrm{s}. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is: (Given : speed of light in vacuum \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1} )
 

Option: 1

1.303


Option: 2

1.318


Option: 3

1.13


Option: 4

0.12


Answers (1)

best_answer

\mathrm{\mu _{B}=1.47=\frac{C}{V_{B}}=\frac{3 \times 10^{8}}{V_{B}}}

\mathrm{V_{B}=\frac{3}{1.47} \times 10^{8}}
\mathrm{V_{B}=2.041\times 10^{8}\, m/s\: \rightarrow (1)}

\mathrm{\because \: V_{A}-V_{B}=2.6\times 10^{7}\, m/s\: (Given)}

\mathrm{\therefore \: V_{A}= V_{B}+2.6\times 10^{7}\, m/s\: }
       \mathrm{V_{A}= 2.301\times 10^{8}\, m/s\: }

\mathrm{\mu _{A}= \frac{C}{V_{A}} = 1.304\: \rightarrow (2)}
\mathrm{\frac{\mu _{B}}{\mu _{A}}= \frac{1.47}{1.304} = 1.13}


The correct option is (3)


 

Posted by

Ritika Kankaria

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