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.

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

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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