I think your analogy is not necessarily relevant as viscosity is a property of the fluid whereas refractive index relates to the material properties, i.e. it would describe the pipe in your analogy. 

Refractive index can be directly related to the relative permittivity of a medium, or alternatively its dielectric constant. The dielectric constant characterises how well electric fields propagate through the material, and given that light is an electromagnetic wave, describes how light will propagate through it. 

But Alex, my analogy is not my question. It may be wrong . 

i am asking the reason behind the light prefers to stay in high refractive index region. If you can could you please elaborate this please 

@ Waseem

First have a look at the actual physics and then the analogy.

Physics: Light or in general say EMW comes in small packets (quanta) of energy called photons. As the photon goes through the material, it interacts with the constituent atoms of the material. Depending on the properties of incident radiation and the material, it could give rise to one or many phenomenon such as absorption, attenuation, dispersion, scattering etc.

For the sake of simplicity lets assume the medium is electrically neutral. That is, the dipole moment of all constituent atoms is zero. However, in most medium the incoming radiation field momentarily polarizes the constituent atoms of the material.  As a result the positive charges (protons) and negative charges (electrons) inside the atom are separated by a finite distance giving rise to a non-zero dipole moment of the atom. But, remember this happens at a very short time scale (10^-12 to 10^-18 seconds or so).

Also, note that prior to the disturbance created by the radiation field, the charges (positive and negative) weren't separated by any "appreciable" distance and that's why the dipole moment was zero.

Once the polarization (though at very short time scale due to absorption of the photon) has taken place, each atomic dipole in the material acts as a phased array antenna. But, this polarized state is highly unstable and so the atomic dipole re-emits the absorbed photon with some delay of-course (10^-12 to 10^-18 seconds). After emitting the photon, the atom returns back to its initial de-excited neutral state. The process repeats across the material along the path-length of the photon as it encounters one atom after another along its flight. In essence, the time duration to cover the path inside the material becomes longer compared to the same length traveled in vacuum where there was none to interrupt the photon path. This is why light has different speed in different materials. 

But in reality light's speed  remains the same independent of the material. The speed of light remains the same in going from one atom to another in any material. It is the absorption and re-emission of the photon by the atom which introduces a phase delay in the path of photon. Hence, it is more suitable to refer to "phase speed" of light in any given medium.

Analogy: Think of it as a relay race where runners of a team carrying batons travel with the same speed (speed of light, c). But, when the baton is passed from the hands of one runner to another, some time is elapsed. It introduces a delay! Passing of baton from one runner to another is similar to absorption and re-emission process of photon inside the material. This causes the final person (i.e., the emergent ray) to arrive to the finishing line much later than he would have arrived had there been no delay (case of vacuum).

Hope that clarifies!

Thanks Vikash ...that was really well explained 

Light is concentrated more in a high refractive index material because higher n means that you can store the EM energy in the form of the potential energy of the polarized atoms or molecules comprising the materials.

However, the statement itself is not necessarily true. For instance, you have the slot effect; if you have two high n media sandwiching a low n medium, the majority of the field will be inside the low index medium due to the continuity of the normal component of the displacement field ( the normal component of D= epsilon * E is equal, so lower epsilon means higher E- see for instance https://www.nature.com/articles/nphoton.2008.131)