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In your research lab, a very thin, flat piece of glass with refractive index 1.20 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength λ0 in vacuum at normal incidence onto the surface of the glass. When λ0= 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm.

Respuesta :

Answer:

(a). The thickness of the glass is 868 nm.

(b). The wavelength is 3472 nm.

Explanation:

Given that,

Refractive index = 1.20

Wavelength = 496 nm

Next wavelength = 386 nm

We need to calculate the thickness of the glass

Using formula for constructive interference

[tex]2nt=(m+\dfrac{1}{2})\lambda[/tex]

Put the value into the formula

In first case,

[tex]2nt=(m+\dfrac{1}{2})496[/tex].....(I)

In second case,

[tex]2nt=(m+1+\dfrac{1}{2})386[/tex]

[tex]2nt=(m+\dfrac{3}{2})386[/tex].....(II)

From equation (I) and (II)

[tex](m+\dfrac{1}{2})496=(m+\dfrac{3}{2})386[/tex]

[tex] 110m=336[/tex]

[tex]m=3.0[/tex]

Put the value of m in equation (I)

[tex]2nt=(2+\dfrac{1}{2})496[/tex]

[tex]t=\dfrac{(3+\dfrac{1}{2})496}{2\times1}[/tex]

[tex]t=868\ nm[/tex]

The thickness of the glass is 868 nm.

(b). We need to calculate the wavelength

Using formula of constructive interference

[tex]2nt=(m+\dfrac{1}{2})\lambda[/tex]

[tex]\lambda=\dfrac{2nt}{(m+\dfrac{1}{2})}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{2\times1\times868}{\dfrac{1}{2}}[/tex]

[tex]\lambda=3472\ nm[/tex]

Hence, (a). The thickness of the glass is 868 nm.

(b). The wavelength is 3472 nm.

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