A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?

Respuesta :

Answer:

[tex]N_T=2086285.67[/tex]

Explanation:

Given;

Thickness of the glass plate, [tex]x=2.95\times 10^{-3}\ m[/tex]

refractive index of the glass plate, [tex]n=1.6[/tex]

wavelength of light source in vacuum, [tex]\lambda=600\times 10^{-9}\ m[/tex]

distance between the source and the screen, [tex]d=1.25\ m[/tex]

Distance travelled by the light from source to screen in vacuum:

[tex]d_v=d-x[/tex]

[tex]d_v=1.25-0.00295[/tex]

[tex]d_v=1.24705\ m[/tex]

So the no. of wavelengths in the vacuum:

[tex]N=\frac{d_v}{\lambda}[/tex]

[tex]N=\frac{1.24705}{6\times 10^{-7}}[/tex]

[tex]N\approx2.0784\times 10^{6}[/tex]  .......................(1)

Now we find the wavelength of the light wave in the glass:

[tex]n=\frac{\lambda}{\lambda'}[/tex]

where:

[tex]\lambda'=[/tex] wavelength of light in the medium of glass.

[tex]1.6=\frac{600\times 10^{-9}}{\lambda'}[/tex]

[tex]\lambda'=375\times 10^{-9}\ m=375\ nm[/tex]

Now the no. of wavelengths in the glass:

[tex]N'=\frac{x}{\lambda'}[/tex]

[tex]N'=\frac{2.95\times 10^{-3}}{375\times 10^{-9}}[/tex]

[tex]N'=7.8667\times 10^{3}[/tex] ............................(2)

From (1) & (2):

  • total no. of wavelengths are there between the source and the screen:

[tex]N_T=N+N'[/tex]

[tex]N_T=2086285.67[/tex]

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