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प्रश्न
The wavelength of sodium light in air is 589 nm. (a) Find its frequency in air. (b) Find its wavelength in water (refractive index = 1.33). (c) Find its frequency in water. (d) Find its speed in water.
उत्तर
Given
Wavelength of sodium light in air,
\[\lambda_a = 589\text{ nm} = 589 \times {10}^{- 9} m\]
Refractive index of water, μw= 1⋅33
We know that
\[f = \frac{c}{\lambda},\]
\[\text{where c = speed of light} = 3 \times {10}^8 m/s\]
f = frequency
λ = wavelength
(a) Frequency in air,
\[f_{\text{air}} = \frac{c}{\lambda_a}\]
\[f_{\text{air}} = \frac{3 \times {10}^8}{589 \times {10}^{- 9}}\]
\[= 5 . 09 \times {10}^{14}\text{ Hz}\]
(b)
Let wavelength of sodium light in water be \[\lambda_w.\]
We know that
\[\frac{\mu_a}{\mu_w} = \frac{\lambda_\omega}{\lambda_a}\]
where `μ_a` is the refractive index of air which is equal to 1 and `λ_w` is the wavelength of sodium light in water.
\[\Rightarrow \frac{1}{1 . 33} = \frac{\lambda_\omega}{589 \times {10}^{- 9}}\]
\[ \Rightarrow \lambda_\omega = 443\text{ nm}\]
(c) Frequency of light does not change when light travels from one medium to another.
\[\therefore f_\omega = f_a \]
\[= 5 . 09 \times {10}^{14} Hz\]
(d) Let the speed of sodium light in water be \[\nu_\omega\] and speed in air, `nu_a = c.`
Using \[\frac{\mu_a}{\mu_\omega} = \frac{\nu_\omega}{\nu_a},\] we get
\[\nu_\omega = \frac{\mu_a c}{\mu_\omega}\]
\[= \frac{3 \times {10}^8}{\left( 1 . 33 \right)} = 2 . 25 \times {10}^8 m/s\]
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