Question 15: Waves Optics

Question

In a YDSE, the light of wavelength $$\lambda=5000\mathop A\limits^o $$ is used, which emerges in phase from two slits at $$a$$ distance $$d=3 	imes 10^{-7} \mathrm{~m}$$ apart. A transparent sheet of thickness $$t=1.5 	imes 10^{-7} \mathrm{~m}$$ and refractive index $$\mu=1.25$$ 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 $$y$$.

Accepted Answer

On introduction of a transparent sheet,
$$\Delta x=(\mu-1) t$$
if the central maxima occupies position of the nth fringe, then $$(\mu-1) t=n \lambda=d \sin 	heta$$
$$\begin{aligned}
\Rightarrow \quad \sin 	heta & =\frac{(\mu-1)}{d} t \
& =\frac{(1.25-1) 	imes 1.5 	imes 10^{-7}}{3 	imes 10^{-7}}=0.125
\end{aligned}$$
Therefore, central maxima has angular position, $$	heta=\sin ^{-1}(0.125)=7.18^{\circ} \approx 7.2^{\circ}$$
for small angles, $$\sin 	heta \approx 	an 	heta$$
$$\begin{aligned}
	an 	heta & =\frac{y}{D} \
	herefore \quad \frac{y}{D} & =\frac{(\mu-1) t}{d} \Rightarrow y=\frac{D(\mu-1)}{d} t
\end{aligned}$$

Answer

$$7.2, \frac{D(\mu-1) t}{2 d}$$

Answer

$$7.2, \frac{D(\mu-1) t}{d}$$

Answer

$$6.2, \frac{D(\mu-1) t}{d}$$

Answer

$$5.2, \frac{2 D(\mu+1) t}{d}$$

Waves Optics

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