Question 1

A camera used for aerial surveillance has a lens with a 30-cm maximum aperture and a 42-cm focal length.Assume light of 550-nm wavelength is used and that the resolution of the camera is limited solely by diffraction.What is the angular resolution of the camera at maximum aperture?
A)1.6 µrad
B)2.2 µrad
C)3.2 µrad
D)4.5 µrad
E)6.3 µrad

Accepted Answer

The angular resolution of a camera limited solely by diffraction is given by:

θ = 1.22λ / D

where θ is the angular resolution, λ is the wavelength of light used, and D is the diameter of the lens (maximum aperture).

Substituting the given values:

θ = 1.22(550 nm) / 30 cm = 2.22 µrad

Therefore, the best choice is B.

Question 2

A certain astronomical telescope has a diameter of 5.60 m.Considering only the limitation due to diffraction,what is the minimum angle of resolution for this telescope at a wavelength of 620 nm?
A)0.111 µrad
B)0.311 µrad
C)0.270 µrad
D)0.135 µrad
E)0.405 µrad

Accepted Answer

The minimum angle of resolution for a telescope is given by the Rayleigh criterion:

θmin = 1.22 λ/D

where λ is the wavelength of light and D is the diameter of the telescope. Substituting the given values, we get:

θmin = 1.22 × 620 × 10^-9 m/5.6 m = 0.135 µrad

Therefore, the answer is D.

Question 3

The angular magnification of a refracting telescope is 40 x.When the object and final image are both at infinity,the distance between the eyepiece and the objective is 143.5 cm.The telescope is used to view a distant radio tower.The real image of the tower,formed by the objective,is 6.0 mm in height.The focal point of the eyepiece is positioned at the real image.What is the focal length of the objective lens?
A)137 cm
B)138 cm
C)139 cm
D)140 cm
E)141 cm

Accepted Answer

The angular magnification (M) of a telescope is given by the ratio of the focal length of the objective lens (f_o) to the focal length of the eyepiece (f_e), M = f_o / f_e. Given that M = 40x and the distance between the eyepiece and the objective (which is approximately the sum of their focal lengths for distant objects) is 143.5 cm, we can set up the equation: f_o + f_e = 143.5 cm and M = f_o / f_e = 40. Solving these equations simultaneously, we find that f_o = 140 cm.

Question 4

What is the limiting angle of resolution for the eye if the pupil diameter of the eye is 4.0 mm,the wavelength of the light is 600 nm,and index of refraction of the liquid in the eye is 1.34?
A)0.42 mrad
B)0.21 mrad
C)0.14 mrad
D)0.11 mrad
E)0.26 mrad

Accepted Answer

The limiting angle of resolution can be determined as:

θ = 1.22 (λ/D)

where θ is the limiting angle of resolution, λ is the wavelength of the light, and D is the pupil diameter of the eye.

Substituting the given values, we get:

θ = 1.22 (600 nm / 4.0 mm)

θ = 0.14 mrad

Therefore, the limiting angle of resolution for the eye is 0.14 mrad, which is closest to choice C, 0.14 mrad.

Question 5

Treat each of your eyes as a circular aperture of diameter 3.5 mm.Light of wavelength
500 nm is used to view two point sources that are 894 m distant from you.How far apart must these two point sources be if they are to be just resolved by your eye? Assume that the resolution is diffraction limited and use Rayleigh's criterion.

Accepted Answer

The answer of Treat each of your eyes as a...

Question 6

The angular magnification of a refracting telescope is 40 x.When the object and final image are both at infinity,the distance between the eyepiece and the objective is 143.5 cm.The telescope is used to view a distant radio tower.The real image of the tower,formed by the objective,is 6.0 mm in height.The focal point of the eyepiece is positioned at the real image.What is the angle subtended by the final image of the tower.
A)0.15 rad
B)0.17 rad
C)0.19 rad
D)0.21 rad
E)0.23 rad

Accepted Answer

Angular magnification, m = -fₒ/fₑ where fₒ and fₑ are the focal lengths of the objective lens and eyepiece, respectively. We have m = 40.
The distance between the objective and eyepiece is given by d = fₒ + fₑ = 143.5 cm.
Using the thin lens equation, 1/f = 1/d₀ + 1/dᵢ where d₀ and dᵢ are the object and image distances, respectively, we can find the focal length of the objective lens:
1/-fₒ = 1/d₀ - 1/∞
1/fₒ = 1/d₀
Using d₀ = ∞, we get fₒ = ∞, which is expected for a telescope. This implies that the objective forms a real image at its focal point, which is the image distance for the eyepiece.
Also, using the equation for angular magnification, we have:
m = -fₒ/fₑ = -(d/fₑ - d)
Solving for fₑ, we get fₑ = -d/m - d = -143.5/40 - 143.5 = -146.1 cm.
Since the real image of the tower is formed at the focal point of the objective, its height is given by:
hi/ho = -do/di = m
where ho is the height of the object (tower), and di and do are the distances of the object and image from the objective, respectively. We have m = 40 and ho = 6.0 mm. Using do = ∞, we get di = fₒ = ∞. Therefore, hi = ho*m = 240 mm.
Finally, the angle subtended by the final image can be found using:
tan(θ) = hi/fₑ = -hi/d
where θ is the angle subtended by the final image at the eyepiece, and d = dᵢ + fₑ = fₒ = ∞ is the distance between the eyepiece and final image. We have hi = 240 mm and fₑ = -146.1 cm. Therefore, θ = tan⁻¹(-hi/d) = 0.17 rad. Thus, the answer is (B).

Question 7

Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the aperture.The diameter of the central bright area is 5.4 cm.What is the diameter of the aperture?
A)110 μm
B)2.0 μm
C)6600 μm
D)960 μm

Accepted Answer

The diameter of the central bright area (first dark fringe) in a diffraction pattern can be calculated using the formula for the first minimum in a circular aperture diffraction, which is $D = 1.22 \lambda L / d$, where $D$ is the diameter of the central bright spot, $\lambda$ is the wavelength of light, $L$ is the distance to the screen, and $d$ is the diameter of the aperture. Rearranging this formula to solve for $d$, we get $d = 1.22 \lambda L / D$. Substituting the given values ($\lambda = 633 \times 10^{-9}$ m, $L = 4.0$ m, and $D = 5.4 \times 10^{-2}$ m), we find $d \approx 1.10 \times 10^{-4}$ m or 110 μm.

Question 8

A radio telescope 200 m in diameter is used to investigate sources emitting a 21-cm wavelength wave.What is the minimum angular separation of the sources that can be resolved by this system?
A)0.073°
B)0.030°
C)0.0013°
D)0.154°
E)0.0026°

Accepted Answer

The minimum angular resolution of a telescope can be calculated using the formula $\theta = \frac{1.22 \lambda}{D}$, where $\theta$ is the angular resolution in radians, $\lambda$ is the wavelength of the light (or other electromagnetic radiation), and $D$ is the diameter of the telescope's aperture. Given $\lambda = 21$ cm = 0.21 m and $D = 200$ m, substituting these values gives $\theta \approx 1.22 \times \frac{0.21}{200} = 0.00127$ radians. Converting to degrees ($1 \text{ radian} = 57.2958^\circ$), $\theta \approx 0.00127 \times 57.2958 = 0.073^\circ$.

Question 9

The objective and the eyepiece of a refracting astronomical telescope have focal lengths of 320 cm and 4.0 cm,respectively.The telescope is used to view Neptune and the final image is set at infinity.The diameter of Neptune is 4.96 × 10⁷m and the distance from the earth at the time of observation is 4.4 × 10¹²m.What is the angle (in mrad)subtended by the final telescopic image of Neptune? A)0.90 mrad B)1.1 mrad C)1.3 mrad D)1.5 mrad E)1.7 mrad

Accepted Answer

The answer of The objective and the eyepiece of a...

Question 10

A 10-inch telescope (25.4 cm in diameter)is used to determine if what appears to be one star is actually two stars.Stars are so far away that they are essentially point sources.How close (in angle)can the two stars be and still be resolved by this telescope if it is focusing light of wavelength of 550 nm? (Consider only the limitation due to diffraction.) A)4.2 × 10^-8 degree B)2.6 × 10^-6 degree C)3.0 × 10^-4 degree D)1.5 × 10^-4 degree E)6.6 × 10^-8 degree

Accepted Answer

The answer of A 10-inch telescope (25.4 cm in diameter)is...

Question 11

Light of wavelength 500 nm illuminates a round 0.50-mm diameter hole.A screen is placed 6.3 m behind the slit.What is the diameter of the central bright area on the screen?
A)15 mm
B)270 μm
C)7.7 mm
D)3800 μm

Accepted Answer

The answer of Light of wavelength 500 nm illuminates a...

Question 12

If the headlights on a car are separated by 1.3 m,how far down the road can they be resolved if the angular resolution of the eye is 5.0 × 10^-4 rad and the person has excellent vision? A)1.3 km B)5.0 km C)4.8 km D)0.65 km E)2.6 km

Accepted Answer

The answer of If the headlights on a car are...

Quiz 35: Optical Instruments

A camera used for aerial surveillance has a lens with a 30-cm maximum aperture and a 42-cm focal length.Assume light of 550-nm wavelength is used and that the resolution of the camera is limited solely by diffraction.What is the angular resolution of the camera at maximum aperture?

A certain astronomical telescope has a diameter of 5.60 m.Considering only the limitation due to diffraction,what is the minimum angle of resolution for this telescope at a wavelength of 620 nm?

What is the limiting angle of resolution for the eye if the pupil diameter of the eye is 4.0 mm,the wavelength of the light is 600 nm,and index of refraction of the liquid in the eye is 1.34?

Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the aperture.The diameter of the central bright area is 5.4 cm.What is the diameter of the aperture?

A radio telescope 200 m in diameter is used to investigate sources emitting a 21-cm wavelength wave.What is the minimum angular separation of the sources that can be resolved by this system?

Light of wavelength 500 nm illuminates a round 0.50-mm diameter hole.A screen is placed 6.3 m behind the slit.What is the diameter of the central bright area on the screen?

If the headlights on a car are separated by 1.3 m,how far down the road can they be resolved if the angular resolution of the eye is 5.0 × 10-4 rad and the person has excellent vision?

If the headlights on a car are separated by 1.3 m,how far down the road can they be resolved if the angular resolution of the eye is 5.0 × 10^-4 rad and the person has excellent vision?