Question

Two optically flat glass plates, 16.0 cm long, are in contact at one end and separated by 0.0200 mm at the other end. The space between the plates is occupied by oil with index of refraction 1.45. The index of refraction of the glass plates is 1.55. The plates are illuminated at normal incidence with monochromatic light, and fringes are observed. If the dark fringes are spaced 2.00 mm apart, what is the wavelength of the monochromatic lig

Answer 1

Answer:

The wavelength of monochromatic light is - 725nm

Explanation:

Lets calculate -

Given - The dark fringes are  2.0mm wide and the glass plate is 16.0cm

 Thus , the number of fringes is - $n=\frac{16cm}{0.2mm}$

                                  Here , converting the units into metre

                                                 = $\frac{16m}{0.2\times10^-^3m}$

                                                   = 80

Now , the expression used here is - $2nt=m\pi$

        Putting the given values,

                    $2\times (1.45)\times(2.0\times10^-^3)=80\pi$

                     $\pi =\frac{2\times1.45\times2.0\times10^-^3}{80}$

                     $\pi =7.25\times10^-^7m$

                      $=725\times10^-^9m$

                       =725nm

Therefore , the answer is 725nm .                            

Answer 2

The wavelength of the monochromatic is mathematically given as

$\pi =725nm$

Wavelength of the monochromatic

Question Parameters:

Two optically flat glass plates, 16.0 cm long,

The space between the plates is occupied by oil with index of refraction 1.45.

The index of refraction of the glass plates is 1.55.

Generally the equation for the    number of fringes is is mathematically

given as

$n=\frac{16}{0.2}$  convert to m

n=80

Therefore

$2* (1.45)*(2.0*10^-^3)=80\pi\\\\ \pi =7.25*10^-^7m$

$\pi =725nm$

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