Chapter 7, Problem 7.24QAP

Interpretation Introduction

(a)

Interpretation:

A spread sheet should be created to determine the Bragg wavelength for the given information.

Concept introduction:

Couple wave theory will be used to determine the behavior of holographic filters. For a holographic optical element, the Bragg wavelength ${\text{λ}}_{\text{b}}$ is determined by:

${\text{λ}}_{\text{b}}\text{=2ndcosθ}$

Where the grating material’s refractive index is n, grating period is d and the angle of incidence of the radiation beam is given by $\theta$.

Explanation of Solution

Given:

A holographic grating

The grating space is d = 17.1 µm

The refractive index n is =1.53

At angles of incidence from 0-90 degrees.

The spreadsheet for the conversion of $\theta$ from degree to radian circulation of $\cos \theta$ and the calculation of the Bragg Wavelength is given below:

Angle of incidence (θ) in degree		Conversion of θ from degree to radian		Cos θ		Bragg wavelength
0		0		1		52.326
10		0.17453293		0.984807753		51.53105048
20		0.34906585		0.939692621		49.17035608
30		0.52359878		0.866025404		45.31564528
40		0.6981317		0.766044443		40.08404153
50		0.87266463		0.64278761		33.63450446
60		1.04719755		0.5		26.163
70		1.22173048		0.342020143		17.89654602
80		1.3962634		0.173648178		9.086314545
90		1.57079633		6.12574E-17		3.20536E-15
CellD3=RADIANS(B7)
CellF3=COS(D7)
CellH3=2*1.53*17.1*F3

Interpretation Introduction

(b)

Interpretation:

The angle at which the Bragg wavelength is 462 nm should to calculated.

Concept introduction:

Couple wave theory will be used to determine the behavior of holographic filters. For a holographic optical element, the Bragg wavelength ${\text{λ}}_{\text{b}}$ is determined by:

${\text{λ}}_{\text{b}}\text{=2ndcosθ}$

Where the grating material’s refractive index is n, grating period is d and the angle of incidence of the radiation beam is given by $\theta$.

Answer to Problem 7.24QAP

The value of angle of incidence is 89.5°, when the Bragg wavelength ${\lambda }_b$ is 462 nm.

Explanation of Solution

The value of Bragg wavelength is 462 nm i.e. 0.462 µm.

Substituting the value for Bragg wavelength in the as 0.462 µm, the grating space d as 17.1 µm, and the refractive index n as 1.53 in the equation, ${\lambda }_b=2nd\cos \theta$:

$0.462\mu m=2\times \left(1.53\right)\times \left(1.71\mu m\right)\cos \theta$

Rearranging the equation:

$\cos \theta =\frac{0.462\mu m}{2\times \left(1.53\right)\times \left(1.71\mu m\right)}=8.83\times {10}^{-3}$

Interpretation Introduction

(c)

Interpretation:

The reason why the Bragg wavelength is also said to be “playback wavelength” at times should be determined.

Concept introduction:

Couple wave theory will be used to determine the behavior of holographic filters. For a holographic optical element, the Bragg wavelength ${\text{λ}}_{\text{b}}$ is determined by:

${\text{λ}}_{\text{b}}\text{=2ndcosθ}$

Where the grating material’s refractive index is n, grating period is d and the angle of incidence of the radiation beam is given by $\theta$.

Explanation of Solution

A hologram is constructed and reconstructed. Bragg angle is that angle at which a hologram is reconstructed.

The wavelength obtained from the Bragg angle is called the Bragg Wavelength. This wavelength is the playback wavelength.

Interpretation Introduction

(d)

Interpretation:

The historical significance of the Bragg wavelength should be determined.

Concept introduction:

Couple wave theory will be used to determine the behavior of holographic filters. For a holographic optical element, the Bragg wavelength ${\text{λ}}_{\text{b}}$ is determined by:

${\text{λ}}_{\text{b}}\text{=2ndcosθ}$

Where the grating material’s refractive index is n, grating period is d and the angle of incidence of the radiation beam is given by $\theta$.

Answer to Problem 7.24QAP

The Bragg wavelength signifies that the X-rays are waves and thus they have historic importance in changing the concept of X-rays.

Explanation of Solution

The thought that X-rays are waves are the main reason behind the historical significance of the Bragg wavelength. In earlier days it was considered that X-rays are composed up of particles. But then the thought was changed and it was considered and experimented that the X-rays consist of waves.

Interpretation Introduction

(e)

Interpretation:

The angular tuning range of the given filer should be determined.

Concept introduction:

Couple wave theory will be used to determine the behavior of holographic filters. For a holographic optical element, the Bragg wavelength ${\text{λ}}_{\text{b}}$ is determined by:

${\text{λ}}_{\text{b}}\text{=2ndcosθ}$

Where the grating material’s refractive index is n, grating period is d and the angle of incidence of the radiation beam is given by $\theta$.

Answer to Problem 7.24QAP

The angular tuning range of the filer is 0° to 40.8°.

Explanation of Solution

Given:

The Bragg wavelength λ_b is 1550 nm. The grating space d is 535 nm.

The filter is tunable over the wavelength range 2nd to $2d\sqrt{n^2-1}$.

The refractive index is 1.53.

$\text{2nd = 2}\left(\text{1}\text{.53}\right)\left(\text{535 nm}\right)\text{=}\left(\text{1070 nm}\right)\left(\text{1}\text{.53}\right)\text{=1640 nm}$

$\text{2d}\sqrt{{\text{n}}^{\text{2}}\text{-1}}\text{=2}\left(\text{535 nm}\right)\sqrt{{\left(\text{1}\text{.53}\right)}^{\text{2}}\text{-1}}\text{=}\left(\text{1070 nm}\right)\sqrt{\text{1}\text{.34}}\text{=1240 nm}$

Thus, the wavelength range is from 1240 nm to 1640 nm.

Substituting in the equation, $\lambda b=2nd\cos \theta$ ,

$\begin{array}{l}\text{1240nm=2(1}\text{.53)(535 nm)cosθ}\\ \text{cos θ=}\frac{\text{1240 nm}}{\text{2(1}\text{.53)(535 nm)}}\text{=0}\text{.757}\\ {\text{θ = cos}}^{\text{-1}}\text{0}\text{.757 = 40}\text{.8}\end{array}$

Again, substituting the value of λ_b as 1640 nm:

$\begin{array}{l}\text{1640 nm=2(1}\text{.53)(535 nm)cosθ}\\ \text{cos θ=}\frac{\text{1640 nm}}{\text{2(1}\text{.53)(535 nm)}}\text{=1}\text{.00}\\ {\text{θ = cos}}^{\text{-1}}\text{1}\text{.00 = 0}\end{array}$

Interpretation Introduction

(f)

Interpretation:

The Bragg wavelength and the wavelength range of the given filter should be determined.

Concept introduction:

Couple wave theory will be used to determine the behavior of holographic filters. For a holographic optical element, the Bragg wavelength ${\text{λ}}_{\text{b}}$ is determined by:

${\text{λ}}_{\text{b}}\text{=2ndcosθ}$

Where the grating material’s refractive index is n, grating period is d and the angle of incidence of the radiation beam is given by $\theta$.

Answer to Problem 7.24QAP

The wavelength range is 490 nm to 647 nm.

Explanation of Solution

Given:

Grating space is 211.5 nm.

Substituting in the equation $\lambda b=2nd\cos \theta$ ,

${\text{λ}}_{\text{b}}\text{=2(1}\text{.53)(211}\text{.5 nm)cosθ=}\left(\text{647cosθ}\right)\text{ nm}$

Thus, the Bragg wavelength is $\left(647\cos \theta \right)nm$. When $\theta$ is known, the Bragg wavelength can be calculated.

The filter is tunable over the wavelength range 2 nd to $2d\sqrt{n^2-1}$.

The refractive index is 1.53.

The grating space is 211.5 nm.

$\text{2nd=2}\left(\text{1}\text{.53}\right)\left(\text{211}\text{.5 nm}\right)\text{=}\left(\text{423 nm}\right)\left(\text{1}\text{.53}\right)\text{= 647 nm}$

$\text{2d}\sqrt{{\text{n}}^{\text{2}}\text{-1}}\text{=2}\left(\text{211}\text{.5 nm}\right)\sqrt{{\left(\text{1}\text{.53}\right)}^{\text{2}}\text{-1}}\text{=}\left(\text{423 nm}\right)\sqrt{\text{1}\text{.34}}\text{= 490 nm}$

Thus, the wavelength range is from 490 nm to 647 nm.

Interpretation Introduction

(g)

Interpretation:

The applications of tunable holographic filters should be determined.

Concept introduction:

A hologram is an image which appears as three dimensional and it can be seen through bare eye. The technology of making and using of holograms is said to be holography.

Explanation of Solution

In single mode fiber multiplexers, tunable holographic filters are used. The filter is a volume-reflection hologram in dichromated gelatin and showcases high reflection efficiencies with very narrow spectrum of bandwidth. They have been used for position tuned wavelength demultiplexers or channel selection switches in multichannel optical fiber systems. By using an insertion loss between input and output single mode fibers of less than 3 decibel with bandwidth nearly 1% of the wavelength band a device can also be constructed.

Interpretation Introduction

(h)

Interpretation:

The refractive index modulation in a given film should be determined.

Concept introduction:

Couple wave theory will be used to determine the behavior of holographic filters. For a holographic optical element, the Bragg wavelength ${\text{λ}}_{\text{b}}$ is determined by:

${\text{λ}}_{\text{b}}\text{=2ndcosθ}$

Where the grating material’s refractive index is n, grating period is d and the angle of incidence of the radiation beam is given by $\theta$.

Answer to Problem 7.24QAP

The refractive index modulation is $8.3\times {10}^{-3}$.

Explanation of Solution

Given:

Wavelength = 633 nm

Thickness = 38 µm

The refractive index modulation for a given wavelength and thickness is given by the formula:

$△n=\frac{\lambda }{2t}$

Substituting wavelength as 633 nm and the thickness as 38 µm in the above equation:

$\begin{array}{l}△n=\frac{\lambda }{2t}\\ △n=\frac{633nm\times {10}^{-3}\frac{\mu m}{nm}}{2\left(38\mu m\right)}=8.3\times {10}^{-3}\end{array}$

Interpretation Introduction

(i)

Interpretation:

The expression for ideal grating efficiency and its value should be determined.

Concept introduction:

Couple wave theory will be used to determine the behavior of holographic filters. For a holographic optical element, the Bragg wavelength ${\text{λ}}_{\text{b}}$ is determined by:

${\text{λ}}_{\text{b}}\text{=2ndcosθ}$

Where the grating material’s refractive index is n, grating period is d and the angle of incidence of the radiation beam is given by $\theta$.

Explanation of Solution

For the real grating, the refractive index modulation is described by:

$△n=\frac{\lambda {\sin }^{-1}\sqrt{D_e}}{\pi t}$

Where D_e is the grating efficiency.

The grating efficiency is expressed in terms of % or fraction. For real grating, when efficiency is expressed in terms of fraction its value will be less than 1.

For ideal grating, the value will be 1, which is 100%.

Substituting D_e as 1 we get:

$\begin{array}{l}△n=\frac{\lambda {\sin }^{-1}\sqrt{D_e}}{\pi t}\\ △n=\frac{\lambda {\sin }^{-1}\sqrt{1}}{\pi t}=\frac{\lambda {\sin }^{-1}\left(1\right)}{\pi t}=\frac{1.57\lambda }{\pi t}\end{array}$

Now, substituting the value of p we would get:

$△n=\frac{1.57\lambda }{3.14t}\Rightarrow \frac{0.5\lambda }{t}=\frac{\lambda }{2t}$

Interpretation Introduction

(j)

Interpretation:

The efficiency for a given grating should be determined.

Concept introduction:

For the real grating, the refractive index modulation is described by:

$△n=\frac{\lambda {\sin }^{-1}\sqrt{D_e}}{\pi t}$

Where D_e is the grating efficiency.

Answer to Problem 7.24QAP

The grating efficiency is 0.79 or 79 %.

Explanation of Solution

Given:

Thickness = 7.5 $\text{μm}$

Refractive index modulation is 0.030

Wavelength = 633 nm

For the real grating, the refractive index modulation is described by:

$△n=\frac{\lambda {\sin }^{-1}\sqrt{D_e}}{\pi t}$

Where D_e is the grating efficiency.

Substituting thickness as 7.5 µm, wavelength as 633 nm, and refractive index modulation as 0.030:

$\begin{array}{l}\text{Vn=}\frac{{\text{λsin}}^{\text{-1}}\sqrt{{\text{D}}_{\text{e}}}}{\text{πt}}\\ \text{0}\text{.030 =}\frac{\left(\text{633 nm}\right){\text{sin}}^{\text{-1}}\sqrt{{\text{D}}_{\text{e}}}}{\text{3}\text{.14}\left(\text{7}\text{.5 μm}\right)}\\ {\sin }^{-1}\sqrt{D_e}=\frac{\text{0}\text{.030×3}\text{.14×}\left(\text{7}\text{.5 μm}\right)}{\left(\text{633 nm}\right)}\Rightarrow \frac{\text{0}\text{.030×3}\text{.14×7}\text{.5 μm}}{{\text{633nm×10}}^{\text{-3}}\frac{\text{μm}}{\text{nm}}}\text{=1}\text{.1}\end{array}$

Taking sin on both sides:

$\begin{array}{l}\sqrt{D_e}\approx 0.89\\ D_e\approx {\left(0.89\right)}^2\approx 0.79\end{array}$

Interpretation Introduction

(k)

Interpretation:

A recipe for holographic films using gelatin, the fabrication of the films and its chemical process should be determined.

Introduction:

Holographic image or patterns are micro embossed over thin plastic films. These are especially used for packaging purpose. Holographic films for fabricating filters are available commercially but these can also be prepared in laboratory using common chemicals like Gelatin.

Explanation of Solution

The recipe for dichromate gelatin for holographic films is:

The gelatin is dissolved in water at 60°C. The potassium dichromate is added to it, then triphenyl dye is added and the ingredients are mixed properly by which they dissolve completely.

In order to fabricate the film, the above solution is then pipetted out on a leveled glass plate, which is then dried for 24 hours.

Gelatin is a polymer containing alpha amino acids, gelatin, potassium dichromate and triphenyl dye together form a complex.

The dried fabricated film is now exposed to an argon laser beam and the diffraction pattern is then measured.