91Ó°ÊÓ

Constructive interference is at $\mathit{n}\mathit{λ}$

and destructive interference is at$\frac{\left(2n+1\right)\mathbf{λ}}{\mathbf{2}}$.

Where n is an integer.

Let the distance between the first virtual image and the straight side be x

We know, Angle of incidenceand angle of refraction on that straight side must be very small so a concentrated image can be formed

$$\begin{gathered} \sin i≈\tan i=\frac{w}{a} \\ \tan i≈\tan r=\frac{w}{x} \end{gathered}$$

By using law of refraction

$n_{i}sini=n_{2}sinr$

We get

$$\begin{gathered} n_{i}sini=n_{2}sinr \\ \frac{w}{a}=n\frac{w}{x} \end{gathered}$$

Rearranging

$x=na$

According to the equal the final virtual image is on the same horizontal position as the light source i.e.(A is small)

We also see that the two virtual images and the light source from a right angled triangle

We also observe

$$\begin{gathered} \tan A=\frac{{\displaystyle \frac{d}{2}}}{na-a})≈A \\ d=2aA(n-1) \end{gathered}$$

Find the distance between the

$$\begin{gathered} d=2(0.2)(3.5×{10}^{-3})(1.5-1) \\ d=7×{10}^{-4} \end{gathered}$$

We know that

$\mathrm{λ}=\frac{dx}{a+b}$

Solve for x

$x=\frac{500×{10}^{-9}(0.2+2)}{7×{10}^{-4}}=1.57×{10}^{-3}m$

Hence, The distance is$x=1.57×{10}^{-3}m$