1 - Iowa State University



41. What is the angle of refraction when a light ray meets the boundary between two materials perpendicularly?

A) 0

B) [pic]

C) [pic]

D) [pic]

E) It depends on index of refraction

Solution:

When a light ray meets the boundary between two materials perpendicularly, both the angle of incidence and the angle of refraction are 0o. Snell’s law shows this to be true for any combination of the indices of refraction of the two materials.

42. Light is passing through water at angle of incidence 55.0° to the surface. At what angle will it leave into the air? (Index of refraction of water is 1.33)

A) 63.7°

B) 53.2°

C) 45.0°

D) 38.9°

E) It will not leave

Solution:

The critical angle for the total internal refractions is given by the equation:

[pic] [pic] [pic].

43. An object is located 2.6 m in front of a plane mirror. The image formed by the mirror appears to be

A) 1.3 m in front of the mirror

B) on the mirror's surface

C) 1.3 m behind the mirror's surface

D) 2.6 m behind the mirror's surface

E) 2.6 m in front of the mirror

Solution:

For plane mirror: [pic]

44. A single convex spherical mirror produces an image which is

A) Always virtual and upright

B) Always virtual and inverted

C) Always real and upright

D) Always real and inverted

E) Real only if the object distance is less than focal distance.

Solution: From the ray diagram follows: “Always virtual and upright”.

45. A mirror at an amusement park shows an upright image of any person who stands 2.0m in front of it. If the image is three times the person’s height, what is the radius of curvature?

A) 2.0 m

B) 3.0 m

C) 4.0 m

D) 5.0 m

E) 6.0 m

Solution:

We find the image distance from the magnification:

[pic] [pic] which gives [pic]

We find the focal length from

[pic] [pic] which gives [pic]

The radius of the concave mirror is [pic]

46. A negative magnification for a mirror means

A) The image is inverted, and the mirror is concave

B) The image is inverted, and the mirror is convex.

C) The image is inverted, and the mirror may be concave or convex

D) The image is upright, and the mirror is convex

E) The image is upright, and the mirror may be concave or convex

Solution:

[pic] Negative m means that [pic] and [pic] have opposite signs – image is inverted.

For convex mirror image is always upright.

47. A negative magnification for a lens means

A) The image is inverted, virtual and the lens is converging

B) The image is inverted, real and the lens is converging

C) The image is, inverted, virtual and the lens is diverging

D) The image is inverted, real and the lens is diverging

E) The image is upright, real and the lens is diverging

Solution:

[pic] Negative m means that [pic] and [pic] have opposite signs– image is inverted.

For diverging lens image is always upright.

48. A coin with diameter 2.0 cm placed 8.0 cm in front of a converging lens. The focal length of the lens is 12.0 cm. Where is the image of the coin?

A) At the lens

B) At the focal point in front of the lens

C) At the focal point behind the lens

D) 16 cm behind the lens

E) 24 cm in front of the lens

Solution:

Mirror equation: [pic] [pic] di = -24 cm

49. When a light wave enters into a medium of different optical density,

A) Its speed and frequency change

B) Its speed and wavelength change

C) Its frequency and wavelength change

D) Its speed, frequency, and wavelength change

E) Its speed change, but frequency and wavelength remain unchanged

Solution

Wave frequency, f is determent by wave source. Speed, v is a characteristic of the medium, and wavelength[pic].

50. Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen 5.00 m away are 6.5 cm apart near the center of the pattern. Determine the wavelength of the light.

A) [pic]

B) [pic]

C) [pic]

D) [pic]

E) [pic]

Solution

For constructive interference, the path difference is a multiple of the wavelength:

[pic].

We find the location on the screen from

[pic]

For small angles, we have [pic] which gives

[pic]

For adjacent fringes, [pic] so we have

[pic]

[pic]which gives [pic]

51. Monochromatic light falls on a slit that is [pic] wide. If the angle between the first dark fringes on either side of the central maximum is 30.0° (dark fringe to dark fringe), what is the wavelength of the light used?

A) 450 nm

B) 511 nm

C) 585 nm

D) 627 nm

E) 673 nm

Solution:

The angle from the central maximum to the first minimum is [pic]= 30.0° /2 =15.0°

We find the wavelength from

[pic] [pic]

[pic], which gives [pic]

52. In order to obtain a good single slit diffraction pattern, the slit width could be:

A) (/100

B) (/10

C) (

D) 10(

E) 100(

Solution:

[pic]

If[pic], than [pic]and one can observe several dark and bright fringes.

53. A 3500 - lines/cm grating produces a third-order fringe at a 28.0° angle. What wavelength of light is being used?

A) 421 nm

B) 447 nm

C) 502 nm

D) 631 nm

E) 680 nm

Solution:

We find the wavelength from [pic] [pic]

[pic], which gives [pic]

54. The separation between adjacent maxima in a double-slit interference pattern using monochromatic light is

A) Greatest for red light

B) Greatest for green light

C) Greatest for blue light

D) The same for all colors of light

E) Greatest for red light or for blue light depending on the distance between the slits

Solution:

Conditions for constrictive interference: [pic].

For given index m, angle [pic]is increasing with increasing wavelength[pic].

Maximum value of [pic]is for the maximum value of[pic], which is for red light.

55. A soap film (n = 1.35) is illuminated with red light (l = 680 nm). The soap film is surrounded by air (n = 1.00) on all sides and appears red at a certain point when viewed at normal incidence. What is the minimum thickness of the film at that point?

A) 58 nm

B) 126 nm

C) 187 nm

D) 272 nm

E) 440 nm

Solution:

There is a phase shift on the front surface of the film. For constructive interference we have: [pic]. At min. thickness, m = 0. So,

[pic]

56. A ray of light is refracted through three different materials. Rank the materials according to their index of refraction.

A) [pic]

B) [pic]

C) [pic]

D) [pic]

E) [pic]

Solution:

By looking at the direction and the relative amount that the light rays bend at each interface, we can infer the relative sizes of the indices of refraction in the different materials (bends toward normal = smaller n material to larger n material; bends away from normal = larger n material to smaller n material).

From the first material to the second material the ray bends toward the normal, thus n1 < n2. From the second material to the third material the ray bends away from the normal, thus n2 > n3. Careful inspection shows that the ray in the third material does not bend back away from the normal as far as the ray was in the first material, thus and n1 < n3.

Thus, the overall ranking of indices of refraction is: n1 < n3 < n2.

57. What is Brewster’s angle for a diamond submerged in water if the light is hitting the diamond [pic] while traveling in the water (n =1.33)?

A) 33°

B) 41°

C) 52°

D) 61°

E) 72°

Solution:

Because the light is coming from water to diamond, we find the angle from the vertical from

[pic] which gives [pic]

58. A person has a far point of 14 cm. What power glasses would correct this vision if the glasses were placed 2.0 cm from the eye?

A) +2.0

B) -2.0

C) -4.6

D) -6.5

E) -8.3

Solution

With the glasses, an object at infinity would have its image 14 cm from the eye or [pic] from the lens; [pic]

[pic]

59. A small insect is placed 5.0 cm from a [pic] lens. Calculate the angular magnification.

A) 1.2

B) 2.4

C) 4.1

D) 5.0

E) 6.0

Solution:

Magnification is

[pic]

60. An astronomical telescope has an objective with focal length 85 cm and a [pic] eyepiece. What is the total magnification?

A) -15

B) -20

C) -30

D) -35

E) -41

Solution:

We find the focal length of the eyepiece from the power:

[pic]

The magnification of the telescope is given by

[pic]

Record Sheet

You may fill in this sheet with your choices, detach it and take it with you after the exam for comparison with the posted answers

|41 |51 |

|A) 0 |E) 673 nm |

|42 |52 |

|E) It will not leave |D) 10( |

|43 |53 |

|D) 2.6 m behind the mirror's surface |B) 447 nm |

|44 |54 |

|A) Always virtual and upright |A) greatest for red light |

|45 |55 |

|E) 6.0 m |B) 126 nm |

|46 |56 |

|A) The image is inverted, and the mirror |B) [pic] |

|is concave | |

|47 |57 |

|B) The image is inverted, real and the |D) 61° |

|lens is converging | |

|48 |58 |

|E) 24 cm in front of the lens |E) -8.3 |

|49 |59 |

|B) Its speed and wavelength change |D) 5.0 |

|50 |60 |

|C) [pic] |C) -30 |

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