Question 17: Wave 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

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

Wave Optics

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