Is there any material which has same refractive index as the air? If there is then what would happen if light passes through it?
 A: The refractive index of a material differs from the vacuum because of interactions of the light with electrons in the material. So to a first approximation the refractive index is related to the electron density. Air has a low refractive index because it has a low density so it doesn't have many electrons per cubic metre.
There are no solids or liquids with a refractive index the same as air, mainly because there aren't any solids or liquids with a density as low as air. So whether there are any objects with the same refractive index as air depends on what you mean by object. A cloud of a different gas in air will have a very similar refractive index, but does a cloud of gas count as an object?
An object with the same refractive index as air would be invisible because no light would be scattered at the air/object boundary. That's assuming of course that the object is transparent and isn't coloured.
Although it's not really what you asked, you might be interested to read about metamaterials. This is a way of making something that manipulates light to behave as if it is invisible.
A: Please see http://www.ecse.rpi.edu/~schubert/Reprints/2007-Xi-JQ-et-al-%28Nature-Photonics%29-Thin-film-with-low-refractive-index-for-broadband-elimination-of-Fresnel-reflection.pdf (nature photonics VOL 1 MARCH 2007 p. 176 ) - pretty dense substance with a refractive index 1.05, close to that of air; low broadband reflection.
Let me note that if you need a refractive index close to that of air in a narrow band only, you can use materials with an absorption line close to your narrow band.
A: 
Is there any material which has same refractive index as the air?

No there isn't. There are more answers to this question which address that.


If there is then what would happen if light passes through it?

Let's assume that such an object, say a fabric, exists. According to the lens maker's equation, $$\frac1f=\left(\frac{\mu_2}{\mu_1}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right)$$ For air, $\mu_1 \approx 1$ and same goes for our 'object'. $\mu_2 \approx 1$. Plugging these values in the equation, we get $$\frac1f=0 \Rightarrow f=\infty$$ Et voilà we have a plane mirror (focal length of plane mirror is $\infty$)! So essentially your 'object' becomes invisible.
Light won't pass through it but reflects off.
