Consider a signal that has lowest frequency component $F_l$ and highest frequency component $F_h$. According to the theory of bandpass sampling, this signal can be sampled and succesfully recoverd if sampled at a frequency ($F_s$) twice the difference between highest and lowest frequency.
that is, $F_s=2 \cdot \left(F_h-F_l\right)$.
My doubt is , Is there a condition or restriction on $F_h$ and $F_l$ for bandpass sampling?
It's generally not true that a band pass signal can be sampled and recovered without error if $f_s>2B$ is satisfied, where $B=f_h-f_l$ is the signal's bandwidth. This condition is just necessary but not sufficient.
You have to make sure that the aliased spectra do not overlap. This results in the following condition on the sampling frequency:
$$\frac{2f_h}{n+1}<f_s<\frac{2f_l}{n}\tag{1},\quad n=0,1,\ldots$$
If Eq. $(1)$ can only be satisfied for $n=0$, you get the familiar Nyquist sampling theorem where $f_s$ must be greater than twice the highest frequency of the signal: $f_s>2f_h$. The lowest possible sampling frequency is obtained for the largest integer $n$ such that $(1)$ is still satisfied. This maximum value of $n$ is given by
$$n_{max}=\left\lfloor\frac{f_l}{B}\right\rfloor\tag{2}$$
where $B=f_h-f_l$ is the bandwidth. Eq. $(2)$ is obtained from $(1)$ by computing the largest value of $n$ such that the left-most term is still less than the right-most term.
As an example, take a band pass signal with $f_l=10\text{ kHz}$ and $f_h=30\text{ kHz}$. From $(2)$, Eq. $(1)$ can only be satisfied for $n=0$, so you have to choose $f_s>2f_h=60\text{ kHz}$, which is a stricter condition than just requiring $f_s>2B=40\text{ kHz}$. However, if $f_l=10\text{ kHz}$ and $f_h=14\text{ kHz}$, the maximum value of $n$ for which $(1)$ is satisfied is $n=2$, which gives you a range $2\cdot 14/3=9.33<f_s<10$ of possible sampling frequencies, all of which of course satisfy $f_s>2B=8\text{ kHz}$.
I think it might be useful to some to see how the formula Matt posted was derived.
$$\frac{2f_h}{n+1}<f_s<\frac{2f_l}{n}\tag{1},\quad n=0,1,\ldots$$
I was inspired by this website and I am not 100% sure if my proof is rigorous enough but here goes,
When the bandpass signal (with $f_l$ being the lowest and $f _h$ being the highest frequency in it's spectra, it's fourier transform is given at the bottom right of the image above) is sampled at $f_s < 2f_h$, aliasing causes copies to be set up at $nf_s$ as shown in the right top image.
Depending on how large $f_l$ is, a larger multiple of $f_s$ is needed for the mirror image of the original spectra to be shifted enough to overlap.
The edge case is when the mirror image of the original spectra ($-f_h$ to $-f_l$) has a right shifted $n^{th}$ alias ($-f_h+nf_s$ to $-f_l +nf_s$) which manages to just fit inside the gap between the original spectra ($f_l$ to $f_h$) and it's immediate left alias ($f_l-f_s$ to $f_h-f_s$).
So now there are two conditions, one condition such that the left side of the $n^{th}$ alias does not overlap with the immediate left alias,
$$f_h - f_s < -f_h +nf_s$$
$$\frac{2f_h}{n+1} < f_s$$
and the right side of the $n^{th}$ alias does not overlap with the original,
$$-f_l + nf_s < f_l $$ $$f_s < \frac{2f_l}{n}$$
So for the largest value of $n$ such that this inequality is satisfied, we can find the range of sampling frequencies at which the aliases created are mutually exclusive and the original continuous function, or a frequency-shifted version of it, can be recovered.
From the inequality $(1)$,
$$\frac{f_h}{f_l} < 1 + \frac{1}{n}$$
$$\frac{f_h - f_l}{f_l} < \frac{1}{n}$$
$$n < \frac{f_l}{B}$$
Where $B = f_h - f_l$ is the the bandwidth. The largest value $n$ can take is $n_{max}=\left\lfloor\frac{f_l}{B}\right\rfloor$ just as Matt has given.
Figure 6.4.3 of Proakis and Manolakis, Digital Signal Processing, 4th ed. has a nice chart of the allowed and disallowed regions for alias-free bandpass sampling.
