Question

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Answer 1

4) Wedge shape film of index of refraction n_w = 1.33

Light refecting from top surface undergoes phase change $\pi$ , equivalent path difference 0.5$\lambda$

Light reflecting from bottom surface undergoes zero phase and path difference

Total path difference between top and bottom surface = 0.5$\lambda$

Thus condition for maxima or bright fringe

Path difference inside film 2L = (n+(1/2)$\lambda$_n) where, n= 0,1,2,3..... and $\lambda$ _n= wavelength inside film

  $2L = (n + \frac{1}{2})\frac{\lambda }{n_{w}}$..................................................1)

where n_w = 1.33 , refractive index of medium, L = thickness of the film

Condition for minima or dark fringe

$2L = (n) \frac{\lambda }{n_{w}}$ ................................................2)

a) when L = 0

Total Path difference of waves reflecting from top and bottom surface

= 0.5 $\lambda$

Thus waves are out of phase with each other. which produces dark fringe

b) $\lambda$ = 630 nm

At 4^th bright fringe n = 4

Substituting value of $\lambda$ and n in equation 1

$2L = (4 + \frac{1}{2})\frac{630\times 10^{-9} }{1.33}$

  $L = 1.065\mu m$

c) Difference in thickness for adjacent bright and dark fringe

for n =1 ,

  $2(L_{1d}-L_{1b})= \frac{\lambda }{n_{w}}-\frac{\lambda }{2n_{w}}$

$L_{1d}-L_{1b}= \frac{\lambda }{4n_{w}}$

$L_{1d}-L_{1b}= \frac{630\times 10^{-9} }{4\times 1.33}$

$L_{1d}-L_{1b}= 0.118\mu m$