In a YDSE, the light of wavelength λ=5000 Ã is used, which emerges...

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 $y$.

Options

1. A$4.9^{{}^{\circ}}$ and $\frac{D(\mu-1) t}{2 d}$
2. B$4.9^{{}^{\circ}}$ and $\frac{\mathrm{D}(\mu-1) \mathrm{t}}{\mathrm{d}}$
3. C$3.9^{{}^{\circ}}$ and $\frac{D(\mu+1) t}{d}$
4. D$2.9^{{}^{\circ}}$ and $\frac{2 \mathrm{D}(\mu+1) \mathrm{t}}{\mathrm{d}}$

Correct answer

B. $4.9^{{}^{\circ}}$ and $\frac{\mathrm{D}(\mu-1) \mathrm{t}}{\mathrm{d}}$

Step-by-step solution

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$4.9 For small angles, \sin \theta=\theta=\tan \theta \Rightarrow \quad \tan \theta=\frac{y}{D}\therefore \quad \frac{\mathrm{y}}{\mathrm{D}}=\frac{(\mu-1) \mathrm{t}}{\mathrm{d}} \Rightarrow \mathrm{y}=\frac{\mathrm{D}(\mu-1) \mathrm{t}}{\mathrm{d}}$

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