Qstnid -Iv-2-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. (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. - 17: Light Waves

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NEET-XII-Physics

17: Light Waves

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#2
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. (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.
Ans : Given:
Wavelength of sodium light in air, `` {\lambda }_{\,\mathrm{\,a\,}}=589\,\mathrm{\,nm\,}=589\times {10}^{-9}\,\mathrm{\,m\,}``
Refractive index of water, μ_w= 1⋅33
We know that `` f=\frac{c}{\lambda }``,
`` \,\mathrm{\,where\,}c=\,\mathrm{\,speed\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,}=3\times {10}^{8}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
f = frequency
λ = wavelength (a) Frequency in air, `` {f}_{\,\mathrm{\,air\,}}=\frac{c}{{\lambda }_{\,\mathrm{\,a\,}}}``
`` {f}_{\,\mathrm{\,air\,}}=\frac{3\times {10}^{8}}{589\times {10}^{-9}}``
`` =5.09\times {10}^{14}\,\mathrm{\,Hz\,}`` (b) Let wavelength of sodium light in water be `` {\lambda }_{\,\mathrm{\,w\,}}``.
We know that
`` \frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,a\,}}}{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,w\,}}}=\frac{{\lambda }_{\,\mathrm{\,\omega \,}}}{{\lambda }_{\,\mathrm{\,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 }_{\,\mathrm{\,\omega \,}}}{589\times {10}^{-9}}``
`` \Rightarrow {\lambda }_{\,\mathrm{\,\omega \,}}=443\,\mathrm{\,nm\,}`` (c) Frequency of light does not change when light travels from one medium to another.
`` \therefore {f}_{\,\mathrm{\,\omega \,}}={f}_{\,\mathrm{\,a\,}}``
`` =5.09\times {10}^{14}\,\mathrm{\,Hz\,}`` (d) Let the speed of sodium light in water be `` {\nu }_{\,\mathrm{\,\omega \,}}``
and speed in air, v_a = c.
Using `` \frac{{\mu }_{\,\mathrm{\,a\,}}}{{\mu }_{\,\mathrm{\,\omega \,}}}=\frac{{\nu }_{\,\mathrm{\,\omega \,}}}{{\nu }_{\,\mathrm{\,a\,}}}``, we get:
`` {\nu }_{\,\mathrm{\,\omega \,}}=\frac{{\mu }_{\,\mathrm{\,a\,}}c}{{\mu }_{\,\mathrm{\,\omega \,}}}``
`` =\frac{3\times {10}^{8}}{\left(1.33\right)}=2.25\times {10}^{8}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
Page No 380: (a) Frequency in air, `` {f}_{\,\mathrm{\,air\,}}=\frac{c}{{\lambda }_{\,\mathrm{\,a\,}}}``
`` {f}_{\,\mathrm{\,air\,}}=\frac{3\times {10}^{8}}{589\times {10}^{-9}}``
`` =5.09\times {10}^{14}\,\mathrm{\,Hz\,}`` (b) Let wavelength of sodium light in water be `` {\lambda }_{\,\mathrm{\,w\,}}``.
We know that
`` \frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,a\,}}}{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,w\,}}}=\frac{{\lambda }_{\,\mathrm{\,\omega \,}}}{{\lambda }_{\,\mathrm{\,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 }_{\,\mathrm{\,\omega \,}}}{589\times {10}^{-9}}``
`` \Rightarrow {\lambda }_{\,\mathrm{\,\omega \,}}=443\,\mathrm{\,nm\,}`` (c) Frequency of light does not change when light travels from one medium to another.
`` \therefore {f}_{\,\mathrm{\,\omega \,}}={f}_{\,\mathrm{\,a\,}}``
`` =5.09\times {10}^{14}\,\mathrm{\,Hz\,}`` (d) Let the speed of sodium light in water be `` {\nu }_{\,\mathrm{\,\omega \,}}``
and speed in air, v_a = c.
Using `` \frac{{\mu }_{\,\mathrm{\,a\,}}}{{\mu }_{\,\mathrm{\,\omega \,}}}=\frac{{\nu }_{\,\mathrm{\,\omega \,}}}{{\nu }_{\,\mathrm{\,a\,}}}``, we get:
`` {\nu }_{\,\mathrm{\,\omega \,}}=\frac{{\mu }_{\,\mathrm{\,a\,}}c}{{\mu }_{\,\mathrm{\,\omega \,}}}``
`` =\frac{3\times {10}^{8}}{\left(1.33\right)}=2.25\times {10}^{8}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
Page No 380:

#2-a
Find its frequency in air.
Ans : Frequency in air, `` {f}_{\,\mathrm{\,air\,}}=\frac{c}{{\lambda }_{\,\mathrm{\,a\,}}}``
`` {f}_{\,\mathrm{\,air\,}}=\frac{3\times {10}^{8}}{589\times {10}^{-9}}``
`` =5.09\times {10}^{14}\,\mathrm{\,Hz\,}``

#2-b
Find its wavelength in water (refractive index = 1.33).
Ans : Let wavelength of sodium light in water be `` {\lambda }_{\,\mathrm{\,w\,}}``.
We know that
`` \frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,a\,}}}{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,w\,}}}=\frac{{\lambda }_{\,\mathrm{\,\omega \,}}}{{\lambda }_{\,\mathrm{\,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 }_{\,\mathrm{\,\omega \,}}}{589\times {10}^{-9}}``
`` \Rightarrow {\lambda }_{\,\mathrm{\,\omega \,}}=443\,\mathrm{\,nm\,}``

#2-c
Find its frequency in water.
Ans : Frequency of light does not change when light travels from one medium to another.
`` \therefore {f}_{\,\mathrm{\,\omega \,}}={f}_{\,\mathrm{\,a\,}}``
`` =5.09\times {10}^{14}\,\mathrm{\,Hz\,}``

#2-d
Find its speed in water.
Ans : Let the speed of sodium light in water be `` {\nu }_{\,\mathrm{\,\omega \,}}``
and speed in air, v_a = c.
Using `` \frac{{\mu }_{\,\mathrm{\,a\,}}}{{\mu }_{\,\mathrm{\,\omega \,}}}=\frac{{\nu }_{\,\mathrm{\,\omega \,}}}{{\nu }_{\,\mathrm{\,a\,}}}``, we get:
`` {\nu }_{\,\mathrm{\,\omega \,}}=\frac{{\mu }_{\,\mathrm{\,a\,}}c}{{\mu }_{\,\mathrm{\,\omega \,}}}``
`` =\frac{3\times {10}^{8}}{\left(1.33\right)}=2.25\times {10}^{8}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
Page No 380: