Relating bandwidth and wavelength of a carrier
I will try to derive your formula.  The formula you were given is technically not correct, but on a practical level it is.  I will try to explain below.

The formula for the frequency of any wave (sound, water, light, etc.) is $f = \frac{\nu}{\lambda}$, where $\lambda$ is the wavelength and $\nu$ is the wave's velocity.  This makes intuitive sense if you think about it because you will see more peaks the faster the wave travels, and you will see fewer peaks the longer the wavelength is.  In your case, of course, the wave velocity is $c$.

Bandwidth is the highest frequency minus the lowest frequency.

$$
\begin{align} 
B &amp;= f_h - f_l \\
&amp;= \frac{c}{\lambda_l} - \frac{c}{\lambda_h} \\
&amp;= \frac{c}{\lambda - \frac{\Delta\lambda}{2}} - \frac{c}{\lambda + \frac{\Delta\lambda}{2}}
\end{align}
$$
At this point to make the typing a little easier I will call $\frac{\Delta\lambda}{2}$ "$d$".
$$
\begin{align} 
B &amp;= \frac{c}{\lambda - d} - \frac{c}{\lambda + d} \\
&amp;= \frac{c(\lambda + d) - c(\lambda - d)}{(\lambda - d)(\lambda + d)} \\
&amp;= \frac{2cd}{\lambda^2 - d^2} \\
&amp;= \frac{c\Delta\lambda}{\lambda^2 - \frac{\Delta\lambda^2}{4}}
\end{align}
$$
The formula you were given is, as a practical matter, correct, because usually $\Delta\lambda$ is much smaller than $\lambda$.  In your case $\Delta\lambda$ is 1/10 of $\lambda$, which is actually much bigger than any normal situation.  Usually $\Delta\lambda$ is orders of magnitude smaller than $\lambda$.  Even at 1/10, though, $\frac{\Delta\lambda^2}{4}$ is $\frac{\lambda^2}{400}$, which is negligible.  Thus, it is dropped.

I hope this helps.