Optics Ques 4

  1. In a double slit experiment, when a thin film of thickness $t$ having refractive index $\mu$ is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of $t$ is $(\lambda$ is the wavelength of the light used)

(2019 Main, 12 April I)

(a) $\frac{2 \lambda}{(\mu-1)}$

(b) $\frac{\lambda}{2(\mu-1)}$

(c) $\frac{\lambda}{(\mu-1)}$

(d) $\frac{\lambda}{(2 \mu-1)}$

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Answer:

Correct Answer: 4.( c )

Solution:

Formula:

Optical Path Difference

and if fringe pattern shifts by one frings width, then path difference,

$ \Delta=1 \times \lambda=\lambda $

So, from Eqs. (i) and (ii), we get

$ (\mu-1) t=\lambda \Rightarrow t=\frac{\lambda}{\mu-1} $

Alternate Solution

Path difference introduced by the thin film of thickness $t$ and refractive index $\mu$ is given by

$ \Delta=(\mu-1) t $

$\therefore$ Position of the fringe is

$ x=\frac{\Delta D}{d}=\frac{(\mu-1) t D}{d} $

Fringe width of one fringe is given by

$ \beta=\frac{\lambda D}{d} $

Given that $x=\beta$, so from Eqs. (i) and (ii), we get

$ \begin{array}{ll} \Rightarrow & \frac{(\mu-1) t D}{d}=\frac{\lambda D}{d} \\ \Rightarrow & (\mu-1) t=\lambda \text { or } t=\frac{\lambda}{(\mu-1)} \end{array} $



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