Chapter 25, Problem 97P

(a)

To determine

The relationship between $\Delta \theta$, the angular separation of the two sources, and $\beta$, the angular separation of the two images.

Answer to Problem 97P

The relationship between $\Delta \theta$, the angular separation of the two sources, and $\beta$, the angular separation of the two images is $\underset{\_}{\sin \Delta \theta =n\sin \beta }$.

Explanation of Solution

The Snell’s law can be used to relate the angular separation of the twos sources and the angular separation

By Snell’s the ray 1 passes into the eye without deviation.

Apply Snell’s law for the ray 2.

  $n_{\text{air}}\sin \Delta \theta =n\sin \beta$                                                                                                     (I)

Here, $n_{\text{air}}$ is the refractive index of air, and $n$ is the refractive index of the vitreous fluid.

Since the refractive index of air is $1.00$, the equation (I) can be reduced to,

  $\sin \Delta \theta =n\sin \beta$

Conclusion:

Therefore, the relationship between $\Delta \theta$, the angular separation of the two sources, and $\beta$, the angular separation of the two images is $\underset{\_}{\sin \Delta \theta =n\sin \beta }$.

(b)

To determine

To show that the relation $\sin \beta \ge \sin \varphi$ when $\beta \ge \varphi$, is equivalent to the Rayleigh’s criterion $a\sin \Delta \theta \ge 1.22{\lambda }_0$.

Answer to Problem 97P

The relation $\sin \beta \ge \sin \varphi$ when $\beta \ge \varphi$, is equivalent to the Rayleigh’s criterion $a\sin \Delta \theta \ge 1.22{\lambda }_0$.

Explanation of Solution

Write the expression for the Rayleigh’s criterion.

  $a\sin \Delta \theta \ge 1.22{\lambda }_0$                                                                                                    (II)

Here, $a$ is the diameter of the circular aperture.

Solve equation (V) for $\sin \Delta \theta$.

  $\sin \Delta \theta \ge \frac{1.22{\lambda }_0}{a}$                                                                                                      (III)

Divide equation (VI) by $n$.

  $\frac{\sin \Delta \theta }{n}\ge \frac{1.22{\lambda }_0}{an}$                                                                                                     (IV)

Write relationship between $\Delta \theta$, the angular separation of the two sources, and $\beta$, the angular separation of the two images.

  $\sin \Delta \theta =n\sin \beta$                                                                                                     (V)

Solve equation (II) for $\sin \beta$.

  $\sin \beta =\frac{\sin \Delta \theta }{n}$                                                                                                     (VI)

Use equation (VI) in (IV).

  $\sin \beta \ge \frac{1.22{\lambda }_0}{an}$                                                                                                     (VII)

Write the expression for determining the location of the first minimum of the diffraction pattern in the eye.

  $a\sin \varphi =1.22\lambda$                                                                                                    (VIII)

Here, $\lambda$ is the wavelength of light in the vitreous fluid.

Write the expression for the wavelength of the light in the vitreous fluid.

  $\lambda =\frac{{\lambda }_0}{n}$                                                                                                                  (IX)

Here, and ${\lambda }_0$ is the wavelength of light in vacuum.

Use equation (IX) in (VIII) and solve for $\sin \varphi$.

  $\begin{array}{l}a\sin \varphi =1.22\frac{{\lambda }_0}{n}\\ \sin \varphi =\frac{1.22{\lambda }_0}{na}\end{array}$                                                                                                     (X)

Use equation (X) in (VII).

  $\sin \beta \ge \sin \varphi$                                                                                                         (XI)

The deduction of the equation (XI) from (II) shows that the relation $\sin \beta \ge \sin \varphi$ when $\beta \ge \varphi$, is equivalent to the Rayleigh’s criterion $a\sin \Delta \theta \ge 1.22{\lambda }_0$.

Conclusion:

Therefore, the relation $\sin \beta \ge \sin \varphi$ when $\beta \ge \varphi$, is equivalent to the Rayleigh’s criterion $a\sin \Delta \theta \ge 1.22{\lambda }_0$.