Physics, asked by vedang2803, 8 months ago

A parallel beam of sodium light strikes a film of oil floating water. When viewed at an angle of 300 from the normal the 8th dark band is seen. IF R.I. of oli is 1.46, find thickness of film​

Answers

Answered by bhagu357
2

Answer:

sorry to say but I don't know the Answer

Answered by talasilavijaya
0

Answer:

The thickness of the oil film on water is  is 1.86\times 10^{-5}m

Explanation:

Given the order of dark fringe, n=8

Refractive index of oil,  \mu= 1.46

Wavelength of sodium light, \lambda= 5893\times 10^{-9} m

and assuming the angle of reflection is, \theta=30^{o}

In thin film interference, the condition for minimum intensity is given by

n\lambda = 2\mu t cos\theta

substituting the values in the above formula

8\times 5893\times 10^{-9}  = 2\times 1.46 \times t \times cos30^{o}

\implies \frac{8\times 5893\times 10^{-9}}{2\times 1.46}  =  t \times \frac{\sqrt{3} }{2}

Therefore, the thickness is,

t=\frac{8\times 5893\times 10^{-9}}{1.732\times 1.46}\approx1.86\times 10^{-5}m

Therefore, the thickness of the oil film on water is 1.86\times 10^{-5}m.

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