Apparent wavelength received by observer in doppler's effect I have a few doubts about Doppler's effect.

*

*Does the apparent wavelength received by the observer depend upon whether the observer is moving or not?

*$λ_{apparent}=\ \frac{v_{sound}\pm v_{source}}{f}$. For any general case, is this formula correct regarding the apparent wavelength received by the observer?

Edit: (Example where observer is moving to prove that wavelength is independent of observer's velocity)

*

*suppose velocity of sound is $v$ and approaching observer has velocity $u$, then $λ=\frac{\left(v+u\right)}{f'}$ where, $f'=\frac{\left(v+u\right)}{v}f$, so that when we substitute $f'$, the apparent wavelength does come out to be independent of velocity of observer.

Please tell me if I am right in thinking so!
 A: Everything seems to be right except the calculation of $λ$. that will be perceived by the observer is given as
$λ=\frac{V}{f’}$ (all three are relative or apparent i.e. as perceived by the observer)
So instead of $V$, we write $V-U$ (since the observer is moving away the relative velocity will become slower, you’ve written $V+U$) this is the mistake you’ve made.
So no, it won’t be independent of observer’s velocity. I recommend seeing some of animation on Doppler effect for better visualisation.
A: The formula of Doppler Effect goes as follows:$$f_o=\Bigl( \frac{ v\pm v_0}{v\pm v_s}\Bigl).f_s$$
Here ,
$f_o$=frequency  measured by observer
$f_s$=frequency of source
Note:
When observer moves away from source - sign is used in numerator and if observer moves towards source + sign is used in numerator.
When source moves away from observer + sign is used in denominator and if source moves towards observer - sign is used in denominator.
Basically,
O.$\rightarrow$          S.$\rightarrow$ $f_o=\Bigl( \frac{ v+ v_0}{v+v_s}\Bigl).f_s$
O.$\rightarrow$          S.$\leftarrow$ $f_o=\Bigl( \frac{ v+ v_0}{v- v_s}\Bigl).f_s$
O.$\leftarrow$           S.$\rightarrow$ $f_o=\Bigl( \frac{ v- v_0}{v+v_s}\Bigl).f_s$
O.$\leftarrow$           S.$\leftarrow$ $f_o=\Bigl( \frac{ v- v_0}{v-v_s}\Bigl).f_s$
[Arrow indicates direction of velocity of observer(o) and source(s) and written beside it is the formula for that corresponding situation]
As you can see frequency measured by observer depends both on observer's velocity and also on source's velocity. So since speed of sound doesn't change we can say by $v=f\lambda$ that wavelength will also depend on both the velocities of source as well as observer.
By the way the formula you used would be valid for a stationery source i.e A SPECIAL CASE.
