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IPhO 1983, Теория — Задача 3. Optics – Problem III: Prisms

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IPhO 1983 Theoretical Questions · 7 т.

Условие

Two dispersive prisms having apex angles $\hat{A}_1 = 60^{\circ}$ and $\hat{A}_2 = 30^{\circ}$ are glued as in the figure below ( $\hat{C} = 90^{\circ}$ ). The dependences of refraction indexes of the prisms on the wavelength are given by the relations

n_1(\lambda) = a_1 + \frac{b_1}{\lambda^2};

Two glued prisms forming the rectangle ABCD-like shape: triangle ADC with apex angle A1 at A, refractive index n1, and right triangle ABC with angle A2 at C and index n2, right angle at C; an incident ray arrives from the left on face AD.

n_2(\lambda) = a_2 + \frac{b_2}{\lambda^2}

were

$a_1 = 1{,}1$ , $b_1 = 1 \cdot 10^5\ \mathrm{nm}^2$ , $a_2 = 1{,}3$ , $b_2 = 5 \cdot 10^4\ \mathrm{nm}^2$ .

a. Determine the wavelength $\lambda_0$ of the incident radiation that pass through the prisms without refraction on $AC$ face at any incident angle; determine the corresponding refraction indexes of the prisms.

b. Draw the ray path in the system of prisms for three different radiations $\lambda_{red}$ , $\lambda_0$ , $\lambda_{violet}$ incident on the system at the same angle.

c. Determine the minimum deviation angle in the system for a ray having the wavelength $\lambda_0$ .

d. Calculate the wavelength of the ray that penetrates and exits the system along directions parallel to DC.

Решение

a.

The ray with the wavelength $\lambda_0$ pass trough the prisms system without refraction on $AC$ face at any angle of incidence if:

n_1(\lambda_0)=n_2(\lambda_0).

Because the dependence of refraction indexes of prisms on wavelength has the form:

n_1(\lambda)=a_1+\frac{b_1}{\lambda^2}\qquad (3.1)

n_2(\lambda)=a_2+\frac{b_2}{\lambda^2}\qquad (3.2)

The relation (3.1) becomes:

a_1+\frac{b_1}{\lambda_0^2}=a_2+\frac{b_2}{\lambda_0^2}.\qquad (3.3)

The wavelength $\lambda_0$ has correspondingly the form:

\lambda_0=\sqrt{\frac{b_1-b_2}{a_2-a_1}}.\qquad (3.4)

Substituting the furnished numerical values

\lambda_0=500\,\mathrm{nm}.\qquad (3.5)

The corresponding common value of indexes of refraction of prisms for the radiation with the wavelength $\lambda_0$ is:

n_1(\lambda_0)=n_2(\lambda_0)=1.5.\qquad (3.6)

The relations (3.6) and (3.7) represent the answers of question a.

b.

For the rays with different wavelength $(\lambda_{red},\lambda_0,\lambda_{violet})$ having the same incidence angle on first prism, the paths are illustrated in the figure 1.1.

Prism ray path, Figure 3.1. — Figure 3.1

The draw illustrated in the figure 1.1 represents the answer of question b.

c.

In the figure 1.2 is presented the path of ray with wavelength $\lambda_0$ at minimum deviation (the angle between the direction of incidence of ray and the direction of emerging ray is minimal).

Prism ray path, Figure 3.2. — Figure 3.2

In this situation

n_1(\lambda_0)=n_2(\lambda_0)=\frac{\sin\frac{\delta_{\min}+A'}2}{\sin\frac{A'}2}.\qquad (3.7)

where $m(\hat A')=30^\circ$ , as in the figure 1.1. Substituting in (3.8) the values of refraction indexes the result is

\sin\frac{\delta_{\min}+A'}2=\frac32\sin\frac{A'}2\qquad (3.8)

\delta_{\min}=2\arcsin\left(\frac32\sin\frac{A'}2\right)-\frac{A'}2.\qquad (3.9)

Numerically

\delta_{\min}\simeq30.7^\circ.\qquad (3.10)

The relation (3.11) represents the answer of question c.

d.

Using the figure 1.3 the refraction law on the $AD$ face is

\sin i_1=n_1\sin r_1.\qquad (3.11)

The refraction law on the $AC$ face is

n_1\sin r_1'=n_2\sin r_2.\qquad (3.12)

Prism ray path, Figure 3.3. — Figure 3.3

As it can be seen in the figure 1.3

r_2=A_2\qquad (3.13)

and

i_1=30^\circ.\qquad (3.14)

Also,

r_1+r_1'=A_1.\qquad (3.15)

Substituting (3.16) and (3.14) in (3.13) it results

n_1\sin(A_1-r_1)=n_2\sin A_2\qquad (3.16)

n_1(\sin A_1\cos r_1-\sin r_1\cos A_1)=n_2\sin A_2.\qquad (3.17)

Because of (3.12) and (3.15) it results that

\sin r_1=\frac1{2n_1}\qquad (3.18)

and

\cos r_1=\frac1{2n_1}\sqrt{4n_1^2-1}.\qquad (3.19)

Putting together the last three relations it results

\sqrt{4n_1^2-1}=\frac{2n_2\sin A_2+\cos A_1}{\sin A_1}.\qquad (3.20)

Because $\hat A_1=60^\circ$ and $\hat A_2=30^\circ$ relation (3.21) can be written as

\sqrt{4n_1^2-1}=\frac{2n_2+1}{\sqrt3}\qquad (3.21)

3n_1^2=1+n_2+n_2^2.\qquad (3.22)

Considering the relations (3.1), (3.2) and (3.23) and operating all calculus it results:

\lambda^4(3a_1^2-a_2^2-a_2-1)+(6a_1b_1-b_2-2a_2b_2)\lambda^2+3b_1^2-b_2^2=0.\qquad (3.23)

Solving the equation (3.24) one determine the wavelength $\lambda$ of the ray that enter the prisms system having the direction parallel with $DC$ and emerges the prisms system having the direction again parallel with $DC$ . That is

\lambda=1194\,\mathrm{nm}\qquad (3.24)

\lambda\simeq1.2\,\mu\mathrm m.\qquad (3.25)

The relation (3.26) represents the answer of question d.

Professor Delia DAVIDESCU, National Department of Evaluation and Examination–Ministry of Education and Research- Bucharest, Romania

Professor Adrian S. DAFINEI, PhD, Faculty of Physics – University of Bucharest, Romania

Marking scheme — Question 3

Part	Marking scheme — The theoretical question III — Prisms	Score
III. a.	$n_1(\lambda_0)=n_2(\lambda_0)$	0.50
	$\lambda_0=\sqrt{(b_1-b_2)/(a_2-a_1)}$	0.25
	Final result: $\lambda_0=500\,\mathrm{nm}$	0.25
	$n_1(\lambda_0)=n_2(\lambda_0)=1.5$	0.25
III. a. total		1.25 points
III. b.	For the rays with different wavelength $(\lambda_{red},\lambda_0,\lambda_{violet})$ having the same incidence angle on first prism, the paths are illustrated in the figure III.1.	2.00 points

Part	Marking scheme	Score
III. c.	$n_1(\lambda_0)=n_2(\lambda_0)=\sin\frac{\delta_{\min}+A'}2/\sin\frac{A'}2$	0.25
	$m(\hat A')=30^\circ$	0.25
	$\delta_{\min}=2\arcsin(\frac32\sin\frac{A'}2)-\frac{A'}2$	0.25
	Final result: $\delta_{\min}\simeq30.7^\circ$	0.25
III. c. total		1.00 point
III. d.	The refraction law on the $AD$ face: $\sin i_1=n_1\sin r_1$	0.25

Marking-scheme prism ray diagram III.2. — Figure III.2

Part	Marking scheme	Score
III. d.	The refraction law on the $AC$ face: $n_1\sin r_1'=n_2\sin r_2$	0.25
	$r_1+r_1'=A_1$	0.25
	$r_2=A_2$	0.25
	$i_1=30^\circ$	0.25
	$\sqrt{4n_1^2-1}=(2n_2\sin A_2+\cos A_1)/\sin A_1$	0.50
	$3n_1^2=1+n_2+n_2^2$	0.25
	$\lambda^4(3a_1^2-a_2^2-a_2-1)+(6a_1b_1-b_2-2a_2b_2)\lambda^2+3b_1^2-b_2^2=0$	0.50
	Final result: $\lambda\simeq1.2\,\mu\mathrm m$	0.25
III. d. total		2.75 points
Total score theoretical question III		7.00 points