Question

In YDSE monochromatic light of wavelength$600nm$ incident of slits as shown in figure.

If $S_1{S}_2=3mm$, $OP=11mm$ then

1. If $\alpha =\frac{0.36}{\pi }$ degree then destructive interfaces at point $P$.
2. If $\alpha =\frac{0.36}{\pi }$ degree then constructive interfaces at point $O$.
3. If $\alpha =0$ then constructive interfaces at $O$
4. Fringe width depends on $\alpha$.

Answer 1

Solution:

Step 1: Given information

Wavelength $=$$600nm$

$S_1{S}_2=3mm$

$OP=11mm$

Step 2: Calculation for part (a)

If $\alpha =\frac{0.36}{\pi }$ degree then destructive interfaces at point P.

$\begin{array}{l}\Rightarrow \Delta x=3\times {10}^{-3}\times \frac{0.36}{\pi }\times \frac{\pi }{180}+\frac{3\times 11\times {10}^{-6}}{1}\\ \Rightarrow \Delta x=3900\end{array}$

Now,

$$\begin{gathered} \Rightarrow 3900=\frac{(2n–1)\lambda }{2} \\ \Rightarrow n=7dest \end{gathered}$$

Step 3: Calculation for part (b)

If $\alpha =\frac{0.36}{\pi }$ degree then constructive interfaces at point O.

$$\begin{gathered} \Rightarrow \Delta x=3mm\left(\frac{0.36}{\pi }\right)\left(\frac{\pi }{180}\right) \\ \Rightarrow \Delta x=600nm \end{gathered}$$

$\Rightarrow 600nm=n\times 600nm$

$\Rightarrow n=1const$

Step 4: Calculation for part (c)

If $\alpha =0$ then constructive interfaces at O.

Consider, $\alpha =0,\Delta x=0$

Therefore, constructive interference:

$\Rightarrow \Delta x=d\sin \alpha +d\sin \theta$

$\Rightarrow ∆x=d\alpha +\frac{dy}{D}$

Step 5: Calculation for part (d)

Fringe width depends on $\alpha$.

Therefore, Fringe width does not depend on alpha, $a::b::c$ .