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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?

Answers (1)

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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