A glass of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8 . Light of wavelength `lambda` travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer. It is partly reflected at the upper and the lower surfaces of the layer ant the two reflected rays interface . If `lambda = 648 nm`, obtain the least value of `t ("in" 10^(-8)m )` which the rays interface constructively.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the conditions for constructive interference
For constructive interference to occur, the path difference between the two reflected rays must be an integral multiple of the wavelength. Additionally, we need to account for the phase change that occurs when light reflects off a medium with a higher refractive index.

### Step 2: Calculate the path difference
When light reflects off the upper surface of the thin layer (refractive index 1.8), there is a phase change of λ/2 (half a wavelength). The path difference between the two rays (one reflecting off the upper surface and one reflecting off the lower surface) is given by:
\[ \text{Path difference} = 2 \mu t \]
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