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  • A small transparent slab containing material of μ =1.5 is placed along AS2 (Fig.10.6). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

A small transparent slab containing material of μ =1.5 is placed along AS2 (Fig.10.6). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

Answers (1)

Explanation:-

From the given diagram



For the principal maxima,(path difference is zero)

\\\Delta x=2d \sin \theta +\left [ (\mu -1)L \right ]=0\\ \sin \theta _0=-\frac{L(\mu-1)}{2d}=-\frac{-L(0.5)}{2d}[\therefore L=d/4]\\or\; \; \; \; \; \Rightarrow \sin \theta _0=\frac{-1}{16}

\theta _0 is the angular position corresponding to the principal maxima.

\Rightarrow \; \; \; OP=D\tan \theta _0\approx D \sin \theta _0=\frac{-D}{16}

For the first minima, the path difference is 

\pm \frac{\lambda}{2}

  \\\Delta x=2d\sin \theta _1+0.5L=\pm \frac{\lambda}{2}\\ \sin \theta _1=\frac{\pm \lambda/2-0.5L}{2d}=\frac{\pm \lambda/2-d/8}{2d}\\\\\Rightarrow \; \; \sin \theta _1=\frac{\pm \lambda/2-\lambda/8}{2\lambda}=\pm \frac{1}{4}-\frac{1}{16}

                                          

Posted by

Gurleen Kaur

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