Question

In placing a thin sheet of mica of thickness $12 \times {10^{ - 5}}cm$ in the path of one of the interfering beams in YDSE, the central fringe shifts equal to a fringe width. Find the refractive index of mica. Given $\lambda = 600nm.$
A. $1.5$
B. $1.48$
C. $1.61$
D. $1.56$

Answer 1

Hint: We know that in an interference pattern fringe with is equal to $n\lambda $ and if a thin sheet of refractive index $\mu $ is introduced in the path of one of the interfering beam then there is a shift in interference pattern on the screen and that fringe shift is equal to $(\mu - 1)$ times the width of slit. Since fringe shift is equal to the fringe width we will equate both the expressions to get the required result.

Complete step by step answer:
Given: thickness of mica sheet $t = 12 \times {10^{ - 5}}cm$, Wavelength $\lambda = 600nm$.Let us assume that the fringe shift is equal to $\beta .$Now we know that the fringe shift is given by,
$\beta = (\mu - 1)t$
Given $t = 12 \times {10^{ - 5}}cm$
Therefore we have,
$\beta = (\mu - 1)12 \times {10^{ - 5}}......(1)$
Now we know that the fringe width is given by $n\lambda $
Here, fringe shift is equal to fringe width
i.e. $\beta = n\lambda $
Now since we are dealing with central bright fringe therefore we have $n = 1$
And, we have given $\lambda = 600nm$
Therefore we have,
$
\beta = 1 \times 600 \times {10^{ - 7}} \\
\Rightarrow\beta = 600 \times {10^{ - 7}}......(2) \\ $
From (1) and (2) we have
$
(\mu - 1) \times 12 \times {10^{ - 5}} = 600 \times {10^{ - 7}} \\
\Rightarrow(\mu - 1) = \dfrac{{600 \times {{10}^{ - 7}}}}{{12 \times {{10}^{ - 5}}}} \\
\Rightarrow(\mu - 1) = \dfrac{1}{2} \\
\Rightarrow(\mu - 1) = 0.5 \\
\Rightarrow\mu = 0.5 + 1 \\\
\therefore\mu = 1.5 $
Hence the required refractive index is $1.5$

Hence, option A is correct.

Note: Fringe width is the distance between two successive bright fringes or two successive dark fringes. In the interference pattern, the fringe width is constant for all the fringes. Fringe width is independent of order of fringe. Fringe width is directly proportional to wavelength of the light used. Whereas fringe shift is shift in interference pattern occurred due to introduction of some obstacle in between the beam of light.