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In a YDSE, the light of wavelength $\lambda=5000 Ã$ is used, which emerges in phase from two slits a distance $\mathrm{d}=3 \times 10^{-7} \mathrm{~m}$ apart. Atransparent sheet of thickness $\mathrm{t}=1.5 \times 10^{-7} \mathrm{~m}$ refractive index $\mu=1.17$ is placed over one of the slits. what is the new angular position of the central maxima of the interference pattern, from the centre of the screen? Find the value of $y$.
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Verified Answer
The correct answer is:
$4.9^{\circ}$ and $\frac{\mathrm{D}(\mu-1) \mathrm{t}}{\mathrm{d}}$
The path difference when transparent sheet is introduced $\Delta x=(\mu-1) t$
If the central maxima occupies position of nth fringe, then $(\mu-1) \mathrm{t}=\mathrm{n} \lambda=\mathrm{d} \sin \theta$
$$
\Rightarrow \sin \theta=\frac{(\mu-1) \mathrm{t}}{\mathrm{d}}=\frac{(1.17-1) \times 1.5 \times 10^{-7}}{3 \times 10^{-7}}=0.085
$$
Therefore, angular position of central maxima
$$
\theta=\sin ^{-1}(0.085)=4.88^{\circ} \approx 4.9
$$
For small angles, $\sin \theta=\theta=\tan \theta$
$$
\Rightarrow \quad \tan \theta=\frac{y}{D}
$$
$\therefore \quad \frac{\mathrm{y}}{\mathrm{D}}=\frac{(\mu-1) \mathrm{t}}{\mathrm{d}} \Rightarrow \mathrm{y}=\frac{\mathrm{D}(\mu-1) \mathrm{t}}{\mathrm{d}}$
If the central maxima occupies position of nth fringe, then $(\mu-1) \mathrm{t}=\mathrm{n} \lambda=\mathrm{d} \sin \theta$
$$
\Rightarrow \sin \theta=\frac{(\mu-1) \mathrm{t}}{\mathrm{d}}=\frac{(1.17-1) \times 1.5 \times 10^{-7}}{3 \times 10^{-7}}=0.085
$$
Therefore, angular position of central maxima
$$
\theta=\sin ^{-1}(0.085)=4.88^{\circ} \approx 4.9
$$
For small angles, $\sin \theta=\theta=\tan \theta$
$$
\Rightarrow \quad \tan \theta=\frac{y}{D}
$$
$\therefore \quad \frac{\mathrm{y}}{\mathrm{D}}=\frac{(\mu-1) \mathrm{t}}{\mathrm{d}} \Rightarrow \mathrm{y}=\frac{\mathrm{D}(\mu-1) \mathrm{t}}{\mathrm{d}}$
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