It's thin film interference. Is this a coated sample? You can actually back out the thickness from the fringe period.

http://www.chem.agilent.com/Library/applications/uv90.pdf

I am asking about some in depth physics.. although thanks Marks.

Optical interference (also called Fabry-Pérot interference) IS the in-depth physics behind this, at least if your sample contains one or more thin films. See also https://www.researchgate.net/post/Why_do_I_see_interference_fringes_in_transmittance_spectra_of_ZnO,

or my paper: http://scitation.aip.org/content/aip/journal/jap/117/4/10.1063/1.4905671

I agree with Manuel and would like to add the following:  The condition for constructive interference is 

m* lambda = 2nd  

where d is the film thickness , n is the refractive index , m is an integer. Two consecutive maxima will correspond to two different m and two  different  lambda. We can write:

     lambda*dm + m*d(lambda) =0

and

 d(lambda)  = lambda*dm/m   ~  lambda ( i. e  the peaks broaden with increasing wavelength)

I agree with the earlier comments.  Optical thin film interference is the reason for the ondulation of the transmittance signal, with the film of the left spectra (in the docs file) being thinner than the film on the right spectra (thicker the film, usually more interference extrema you will get).  If you have a single film and a relatively thick layer (as on the right spectra), simple envelop methods (such as Manifacier or Swanepoel) will give you information about the refractive index and thickness of the film.

Thanks all of you for the kind suggestion and the explanations. But yet I have not found any good reason for the peaks broadening with increasing wavelength? Will anyone shed a light on that too????

If you plot the spectrum as a function of 1/wavelength, you may find that the peaks are more eqidistant with similar width.  Look at the équations describing interférence and you will see the dépendance on 1/wavelwngth.