cyfrinairw7pr

2023-02-21

In air, the wavelength of sodium light is 589 nm. (a) Determine its frequency in the air. (b) Find its wavelength in water (refractive index = 1.33). (c) Determine its frequency in water. (d) Determine its speed in water.

Caltolmsn

Beginner2023-02-22Added 9 answers

Given that, for sodium light,

$1=589nm=589\times {10}^{-9}m$(a)${f}_{a}=\frac{3\times {10}^{8}}{589\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}=5.09\times {10}^{14}{\mathrm{sec}}^{-1}$(b)$\frac{\mu a}{\mu w}=\frac{{\lambda}_{w}}{{\lambda}_{q}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{1.33}=\frac{\lambda w}{589\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{w}=443nm$(c)${f}_{w}={f}_{a}\phantom{\rule{0ex}{0ex}}=5.09\times {10}^{14}se{c}^{-1}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{w}=\frac{{\mu}_{a}{v}_{a}}{{\mu}_{w}}\phantom{\rule{0ex}{0ex}}=\frac{3\times 108}{(1.33)}=2.25\times {10}^{8}m/sec.$

$1=589nm=589\times {10}^{-9}m$(a)${f}_{a}=\frac{3\times {10}^{8}}{589\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}=5.09\times {10}^{14}{\mathrm{sec}}^{-1}$(b)$\frac{\mu a}{\mu w}=\frac{{\lambda}_{w}}{{\lambda}_{q}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{1.33}=\frac{\lambda w}{589\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{w}=443nm$(c)${f}_{w}={f}_{a}\phantom{\rule{0ex}{0ex}}=5.09\times {10}^{14}se{c}^{-1}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{w}=\frac{{\mu}_{a}{v}_{a}}{{\mu}_{w}}\phantom{\rule{0ex}{0ex}}=\frac{3\times 108}{(1.33)}=2.25\times {10}^{8}m/sec.$

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