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Question: When a thin wedge-shaped film is illuminated by a parallel beam of light of wavelength 6000 \({A^o}\...

When a thin wedge-shaped film is illuminated by a parallel beam of light of wavelength 6000 Ao{A^o}, 7 fringes are observed in a certain region of the film. What will be the number of fringes observed in the same region of a film if light of wavelength 4200 Ao{A^o} is used?

Explanation

Solution

- Hint: In this question use the relation between refractive index, number of fringes, wavelength of the light and the separation between two surfaces which is given as 2μd=nλ2\mu d = n\lambda so use this concept to reach the solution of the question.

Formula used – 2μd=nλ2\mu d = n\lambda

Complete step by step answer:
As we know that in a thin wedge shaped film there is relation which is given as
2μd=nλ2\mu d = n\lambda .............. (1), where μ\mu = refractive index of the film.
d = separation between two films.
n = number of fringes.
λ\lambda = wavelength of light.
Wavelength of the first beam of light is λ1=6000A0{\lambda _1} = 6000{A^0} and the number of fringesn1=7{n_1} = 7.
And Wavelength of the second beam of light is λ2=4200A0{\lambda _2} = 4200{A^0} and the number of fringesn2={n_2} = ?
Now it is given that the medium is the same in both the cases.
So from equation (1) we can say that
nλ=\Rightarrow n\lambda = Constant.
Therefore it is written as
n1λ1=n2λ2\Rightarrow {n_1}{\lambda _1} = {n_2}{\lambda _2}
Now substitute the values in this equation we have,
7(6000)=n2(4200)\Rightarrow 7\left( {6000} \right) = {n_2}\left( {4200} \right)
Now simplify this we have,
n2=7(6000)4200=10\Rightarrow {n_2} = \dfrac{{7\left( {6000} \right)}}{{4200}} = 10
So there are 10 fringes when the wavelength of light is 4200 Ao{A^o}.
So this is the required answer.
Hence option (B) is the correct answer.

Note – Whenever we face such types of questions the key concept is the formula used in this which is stated above now as we know that refractive index is different for different medium but in this case medium is same so there is no use of refractive index as this is canceled out. So for two cases whose medium is same the formula becomes n1λ1=n2λ2{n_1}{\lambda _1} = {n_2}{\lambda _2} so just simplify substitute the values in this equation and simplify, we will get the required answer.