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Question: A mica sheet of thickness \(1.964\mu m\) and the refractive index \(1.6\) is placed in the path of o...

A mica sheet of thickness 1.964μm1.964\mu m and the refractive index 1.61.6 is placed in the path of one of the interfering waves. Then the mica sheet is removed and the distance between the slits and screen is doubled. If this states the distance between two consecutive maxima or minima is equal to the displacement of fringe pattern on placing a mica sheet, the wavelength of the monochromatic light used in the experiment will be,
A.5892A0  B.5269A0  C.6271A0  D.3875A0  \begin{aligned} & A.5892\overset{0}{\mathop{A}}\, \\\ & B.5269\overset{0}{\mathop{A}}\, \\\ & C.6271\overset{0}{\mathop{A}}\, \\\ & D.3875\overset{0}{\mathop{A}}\, \\\ \end{aligned}

Explanation

Solution

here it is mentioned that the fringe width is equal to the shift of the pattern. The fringe width can be calculated by taking the ratio of the product of the wavelength, order of the fringe and the distance between the slit and screen to the slit width. And the displacement of the fringe pattern can be calculated by taking the ratio of the product of refractive index minus one, thickness of the mica sheet and the distance between the slit and the screen to the slit width. Calculate the wavelength using this relation.

Complete answer:
It is mentioned in the question that the fringe width is equal to the shift in the fringe pattern. The fringe width is given as,
β=λDd\beta =\dfrac{\lambda D}{d}
Where λ\lambda be the wavelength, DDbe the distance between the slit and screen and dd be the slit width. The distance between the screen and slit is doubled.
Substituting the values given in the question gives,
β=λ×2Dd\beta =\dfrac{\lambda \times 2D}{d}
Shift is given by the equation,
s=(μ−1)tDds=\dfrac{\left( \mu -1 \right)tD}{d}
Where μ\mu be the refractive index given as,
μ=1.6\mu =1.6
And tt be the thickness of the sheet given as,
t=1.964×10−6mt=1.964\times {{10}^{-6}}m
Substituting this in the equation will give,
s=(1.6−1)×1.964×10−6×Dds=\dfrac{\left( 1.6-1 \right)\times 1.964\times {{10}^{-6}}\times D}{d}
Now let us equate both the equations as they are equal.
2λDd=(1.6−1)×1.964×10−6×Dd\dfrac{2\lambda D}{d}=\dfrac{\left( 1.6-1 \right)\times 1.964\times {{10}^{-6}}\times D}{d}
Cancelling the common terms and rearranging will give,
λ=(1.6−1)×1.964×10−62\lambda =\dfrac{\left( 1.6-1 \right)\times 1.964\times {{10}^{-6}}}{2}
Therefore the value of the wavelength used is obtained as,
λ=0.5892×10−6=5892A0 \lambda =0.5892\times {{10}^{-6}}=5892\overset{0}{\mathop{A}}\,

So, the correct answer is “Option A”.

Note:
Young's double-slit experiment is done in the question. In which the monochromatic light passes through two narrow slits. This will illuminate a distant screen. A particular pattern of bright and dark fringes is observed in the screen. This interference pattern is formed by the superposition of overlapping waves coming from the two slits.