AC = CO = D, S1 C = S2 C = d

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AC = CO = D, S1 C = S2 C = d << D A small transparent slab containing material of µ =1.5 is placed along AS2. 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?

CBSE | Class 12 | ncert exemplar | Physics | Wave Optics

AC = CO = D, S1 C = S2 C = d << D A small transparent slab containing material of µ =1.5 is placed along AS2. 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?

According to the question,

$\begin{array}{l} \Delta x=2 d \sin \theta+(\mu-1) L \\ \sin \theta 0=-1 / 16 \end{array}$
From central maxima, $O P=-D / 16$
$\sin \theta_{1}=\frac{\pm \lambda / 2-d / 8}{2 d}$
On the positive side,
$\sin \theta_{1}^{+}=3 / 16$
On the negative side,
$\sin \theta_{1}^{-}=-5 / 16$
On the positive side, the first main maximum is at a distance, which is above point. $O$
$D \tan \theta_{1}^{-}=\frac{3 D}{\sqrt{247}}$
On the negative side, the first main minima is at a distance, which is below point. $\mathrm{O}$ $\operatorname{Dtan} \theta_{1}^{-}=\frac{5 D}{\sqrt{16^{2}-5^{2}}}=\frac{5 D}{\sqrt{231}}$