Question

Two light rays, initially in phase and with a wavelength of 500 nm, go through different paths by reflecting from the various mirrors shown in Fig. 35-49. (Such a reflection does not itself produce a phase shift.) (a) What least value of distance will put the rays exactly out of phase when they emerge from the region? (Ignore the slight tilt of the path for ray 2.) (b) Repeat the question assuming that the entire apparatus is immersed in a protein solution with an index of refraction of 1.38.

Short answer

(a) The least distance of d is $50\text{nm}$.

(b) The least value of the distance if the apparatus is immersed in protein solution is $36.2\text{nm}$

Step-by-step solution

Given data

Wavelength of light $\lambda =500 \text{nm}$

Concept used

A light ray is a line (straight or curved) that is perpendicular to the light's wave fronts; its tangent is collinear with the wave vector. Light rays in homogeneous media are straight. They bend at the interface between two dissimilar media and may be curved in a medium in which the refractive index changes

(a) Determine the least value of distance 

Distance is the total path travelled from initial point to final point.

Calculate the least value of d to put the rays exactly out of the phase when they emerge for the region.

Ray 1 travelled a total distance of 7d

Ray 2 travelled a total distance of 2d

The path difference between ray 1 and ray 2 is

$7d-2d=5d$

For the rays to be exactly out of phase this path different must be equal $\frac{\lambda }{2}$, that is,

$5d=\frac{\lambda }{2}$

Here, d is the least distance and $\lambda$ is the wavelength of the light.

Rearrange the above equation for d.

$d=\frac{\lambda }{10}$

Substitute 500 nm for $\lambda$in above equation.

$d=\frac{\left(500\text{nm}\right)}{\left(10\right)}$

Therefore, the least distance of d is 500 nm.

(b) Determine the least value of distance

Calculate least value of the, d to put the rays exactly out of the phase when they emerge from the region if the apparatus is immersed in protein solution.

When the apparatus is immersed in protein solution with a reactive index of n,

Then the path difference becomes $n\left(5d\right)$

For the rays to be exactly out of phase his path difference must be equal to $\frac{\lambda }{2}$, that is,

$n\left(5d\right)=\frac{\lambda }{2}$

Here, n is the reactive index of the protein solution d is the least distance and $\lambda$is the wavelength of the light.

Rearrange the above equation for d.

$d=\frac{\lambda }{n\left(10\right)}$

Substitute 1.38 for n and 500 nm for $\lambda$ in above equation.

$\begin{array}{rcl}d& =& \frac{\left(500nm\right)}{\left(10\right)\left(1.38\right)}\\ & =& 36.2nm\end{array}$

Therefore, the least value of the distance if the apparatus is immersed in protein solution is $36.2nm$.