Question

In Fig.35-51a , the waves along rays 1 and 2 are initially in phase, with the same wavelength $\mathit{\lambda }$in air. Ray 2 goes through a material with length and index of refraction n. The rays are then reflected by mirrors to a common point on a screen. Suppose that we can vary n from $\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}$ to $\mathit{n}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}$. Suppose also that, from $\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}$ to ${\mathit{n}}_{\mathbf{1}}\mathbf{-}{\mathit{n}}_{\mathbf{s}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}$, the intensity I of the light at point P varies with n as given in Fig.35-51b . At what values of n greater than 1.4 is intensity I (a) maximum and (b) zero? (c) What multiple of $\lambda$ gives the phase difference between the rage at point p when$\mathit{n}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}$

Short answer

(a) 1.8.

(b) 1.

(c)$1.25\lambda$

Step-by-step solution

Concept of interference fringes

The alternating bright and the dark band formed due to interferenceis called fringe. When two light waves superimpose it forms constructive interference and destructive interference. The bright band is due to constructive interference and the dark band is due to destructive interference.

(a) Determine the refractive index for maximum intensity

From the graph when $n=1$ intensity is maximum and at $n=1.4$ the intensity is minimum.

Therefore difference in the index of refraction for successive maximum intensity and minimum intensity is $\Delta n=0.4$

So the next maximum intensity at $n=1.4+0.4=1.8$

Therefore, the next maxima are 1.8.

(b) Determine the refractive index for zero intensity

Next minimum will occur at

$n=1.8+0.4=2.2$

Here,$n_1=n=2$ and $n_2=1$

$\begin{array}{rcl}\Delta n& =& n_1-n_2\\ & =& 2-1\\ & =& 1\end{array}$

But $\Delta n=0.4$ gives minimum interference $\Delta n=0.4$

Corresponds to a phase difference of $\frac{\lambda }{2}$

Therefore $\Delta n=1$

(c) Determine the phase difference of wave

When $n=2$

Here,

$\begin{array}{rcl}\Delta n& =& 2-1\\ & =& 1\end{array}$

$\begin{array}{rcl}\text{phase} \text{difference}& =& \frac{\left(\frac{\lambda }{2}\right)}{4}\\ & =& \frac{\lambda }{0.8}\\ & =& 1.25\lambda \end{array}$

At point p when $n=2$ the phase difference between the two rays will $1.25\lambda$

Therefore, the phase difference between the two raysis $1.25\lambda$