Question

You are standing in air and are looking at a flat piece of glass (n = 1.52) on which there is a layer of transparent plastic (n = 1.61). Light whose wavelength is 595 nm in vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark.

Find the smallest possible nonzero value for the thickness of the layer.

Answer 1

Answer:

184.78 nm

Explanation:

The condition for dark fringes in thin film surrounded by air is given by the following relation:

2nt = mλ

The order number of the dark fringe is m=1.

Calculate the smallest possible non zero value for the thickness of the layer is,

2nt = mλ

t = mλ / 2n

= (1) (595 * 10⁻⁹m) / (2) (1.61)

= 1.8478 * 10⁻⁷ m

= 184.78 nm

Answer 2

"184.78 nm" would be the smallest possible nonzero value for the thickness of the layer. A further explanation is below.

Given:

Wavelength,

• $\lambda = 595 \ nm$

Order no. of dark fringes,

• $m = 1$

and,

• $n = 1.52$
• $n = 1.61$

Now,

By using the relation, we get

→ $2nt = m \lambda$

or,

→ $t = \frac{m \lambda}{2n}$

By substituting the values, we get

→    $= \frac{1\times 595\times 10^{-9}}{2\times 1.61}$

→    $= 1.8478\times 10^{-7}$

→    $= 184.78 \ nm$

Thus the above response is right.

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