Question

When a thin wedge-shaped film is illuminated by a parallel beam of light of wavelength 6000 ${A^o}$, 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 ${A^o}$ is used?

Answer 1

- 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 $2\mu d = n\lambda $ so use this concept to reach the solution of the question.

Formula used – $2\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
$2\mu d = n\lambda $.............. (1), where $\mu $= refractive index of the film.
                                                         d = separation between two films.
                                                         n = number of fringes.
                                                        $\lambda $ = wavelength of light.
Wavelength of the first beam of light is ${\lambda _1} = 6000{A^0}$ and the number of fringes${n_1} = 7$.
And Wavelength of the second beam of light is ${\lambda _2} = 4200{A^0}$ and the number of fringes${n_2} = $?
Now it is given that the medium is the same in both the cases.
So from equation (1) we can say that
$ \Rightarrow n\lambda = $ Constant.
Therefore it is written as
$ \Rightarrow {n_1}{\lambda _1} = {n_2}{\lambda _2}$
Now substitute the values in this equation we have,
$ \Rightarrow 7\left( {6000} \right) = {n_2}\left( {4200} \right)$
Now simplify this we have,
$ \Rightarrow {n_2} = \dfrac{{7\left( {6000} \right)}}{{4200}} = 10$
So there are 10 fringes when the wavelength of light is 4200 ${A^o}$.
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 ${n_1}{\lambda _1} = {n_2}{\lambda _2}$ so just simplify substitute the values in this equation and simplify, we will get the required answer.