Question

In newtons ring experiment the
diameter of nine th bright ring
changes from 0.3 cm to 0.26 cm
When a liquid is introduced between
the lens and the glass plate. what will
be the refractive index of the liquid.​

Answer 1

Answer:

Research on the latest instruments used in preparing new elements in the laboratory. What were the instruments used in preparing the newest four elements, nihonium, moscovium, tennessine, and oganesson?

Answer 2

Given,

The diameter of the ring in presence of air ($D_{air}$) is 0.3 cm.

The diameter of the ring in presence of liquid ($D_{liquid}$) is 0.26 cm.

To find,

The refractive index of the liquid (μ).

Solution,

The diameter of the ninth ring in presence of air is 0.3 cm.

The diameter of the ninth ring in presence of liquid is 0.26 cm.

So, the refractive index is the square of the ratio of the diameter in presence of air to the diameter in presence of liquid.

μ= 1.33