Question

A thin plastic sheet of refractive index 1.6 is used to cover one of the slits of a double slit arrangement. The central point on the screen is now occupied by what would have been the 7th bright fringe before the plastic was used. If the wavelength of light is 600nm, what is the thickness (in $\mu $) of the plastic sheet?
(A). 7
(B). 4
(C). 8
(D). 6

Answer 1

Hint – In this question since the central point on the screen is now occupied, by what would have been the 7th bright fringe before the plastic was used therefore n = 7. Use the concept that the change in path length is given as the product of one less than refractive index multiplied by the thickness of the plastic sheet that is changed in path length is $\left( {\mu - 1} \right)t$. This will help to approach the problem.

Formula used: $\left( {\mu - 1} \right)t$.

Complete step-by-step solution -
Given data:
Refractive index of the thin plastic sheet = 1.6
Often denoted by $\mu $
Therefore, $\mu $ = 1.6
Is used to cover one of the slits in a double slit arrangement.
Now the central point on the screen is now occupied by what would have been the 7th bright fringe before the plastic was used.
Therefore, n = 7
Now if the wavelength of the light is 600 nm, often denoted by \[\lambda \],
Therefore, \[\lambda \] = 600 nm = 600$ \times {10^{ - 9}}$m
So we have to calculate the thickness of the plastic sheet in micrometer ($\mu m$)
Now the change in path length is given as the product of one less than refractive index multiplied by the thickness of the plastic sheet.
Therefore, change in path length = $\left( {\mu - 1} \right)t$.................. (1)
Now this change in path length changes the position of the seventh bright fringe which is the product of the fringe number and the wavelength of the light.
Therefore, it is equal to $n\lambda $......................... (2)
Now from equation (1) and (2) we have,
$ \Rightarrow \left( {\mu - 1} \right)t = n\lambda $
Now substitute the values in the above equation we have,
$ \Rightarrow \left( {1.6 - 1} \right)t = 7\left( {600 \times {{10}^{ - 9}}} \right)$
Now simplify this we have,
$ \Rightarrow t = \dfrac{{7\left( {600 \times {{10}^{ - 9}}} \right)}}{{0.6}} = 7000 \times {10^{ - 9}} = 7 \times {10^{ - 6}}$m
Now as we all know $1\mu m = 1 \times {10^{ - 6}}m$
$ \Rightarrow t = 7\mu m$
So this is the required thickness of the sheet.
Hence option (A) is the correct answer.

Note – Young’s double slit experiment is generally performed to resemble that light can depict both the basic nature of waves as well as particles. This shows the duality of the light. The photons that are composed in the light helps depict the particle nature of the light.