Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 25, Problem 97P

(a)

To determine

The relationship between Δθ, the angular separation of the two sources, and β, the angular separation of the two images.

(a)

Expert Solution
Check Mark

Answer to Problem 97P

The relationship between Δθ, the angular separation of the two sources, and β, the angular separation of the two images is sinΔθ=nsinβ_.

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.

  nairsinΔθ=nsinβ                                                                                                     (I)

Here, nair 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,

  sinΔθ=nsinβ

Conclusion:

Therefore, the relationship between Δθ, the angular separation of the two sources, and β, the angular separation of the two images is sinΔθ=nsinβ_.

(b)

To determine

To show that the relation sinβsinϕ when βϕ, is equivalent to the Rayleigh’s criterion asinΔθ1.22λ0.

(b)

Expert Solution
Check Mark

Answer to Problem 97P

The relation sinβsinϕ when βϕ, is equivalent to the Rayleigh’s criterion asinΔθ1.22λ0.

Explanation of Solution

Write the expression for the Rayleigh’s criterion.

  asinΔθ1.22λ0                                                                                                    (II)

Here, a is the diameter of the circular aperture.

Solve equation (V) for sinΔθ.

  sinΔθ1.22λ0a                                                                                                      (III)

Divide equation (VI) by n.

  sinΔθn1.22λ0an                                                                                                     (IV)

Write relationship between Δθ, the angular separation of the two sources, and β, the angular separation of the two images.

  sinΔθ=nsinβ                                                                                                     (V)

Solve equation (II) for sinβ.

  sinβ=sinΔθn                                                                                                     (VI)

Use equation (VI) in (IV).

  sinβ1.22λ0an                                                                                                     (VII)

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

  asinϕ=1.22λ                                                                                                    (VIII)

Here, λ is the wavelength of light in the vitreous fluid.

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

  λ=λ0n                                                                                                                  (IX)

Here, and λ0 is the wavelength of light in vacuum.

Use equation (IX) in (VIII) and solve for sinϕ.

  asinϕ=1.22λ0nsinϕ=1.22λ0na                                                                                                     (X)

Use equation (X) in (VII).

  sinβsinϕ                                                                                                         (XI)

The deduction of the equation (XI) from (II) shows that the relation sinβsinϕ when βϕ, is equivalent to the Rayleigh’s criterion asinΔθ1.22λ0.

Conclusion:

Therefore, the relation sinβsinϕ when βϕ, is equivalent to the Rayleigh’s criterion asinΔθ1.22λ0.

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