Question #102880
A wedge shaped film has refractive index 1.34 and thickness of extreme sides 0 (zero) and t. If a light of wavelength 492 nm is incident normally on it and 20 fringes are obtained, determine t.
1
Expert's answer
2020-04-05T14:49:39-0400

As per the given data in the question,

Refractive index of the film=1.34

thickness of the extreme side 0 and t

wavelength of the light (λ)=492nm(\lambda)=492nm

Number of fringes=20

As per the rule of x-ray reflection on the single layer

2tn2sin2iλ2=mλ2t\sqrt{n^2−\sin^2i}-\dfrac{\lambda}{2}​=mλ

n=refractive index of the medium

Light is getting incident normally, so i=0i=0

So,

2tnλ2=mλ2tn-\dfrac{\lambda}{2}=m\lambda


t=mλ+λ22n\Rightarrow t=\dfrac{m\lambda+\dfrac{\lambda}{2}}{2n}


t=20×492×109+246×1092×1.34\Rightarrow t=\dfrac{20\times492\times10^{-9}+246\times 10^{-9}}{2\times 1.34}


t=3763×109m\Rightarrow t=3763\times10^{-9}m


t=3.76μmt=3.76\mu m



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Comments

Assignment Expert
05.04.20, 21:50

Dear visitor, thank you for your comment. Please check updated solution

Unknown
03.04.20, 13:15

Final answer is wrong , Calculation mistake !

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