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CSEC (CXC) Geometric Optics — practice questions — page 2

Explore 60 free Geometric Optics questions in CSEC (CXC) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Geometric Optics questions and exercises with answers

Showing questions 31–60 of 60.

What is a virtual image?Read the answer and worked solution →In the magnification equation for a thin lens, what does a negative value of m mean?Read the answer and worked solution →A clear glass light bulb sits 0.750 m from a convex lens whose focal length is 0.500 m. What are the image distance and the magnification?Read the answer and worked solution →To use a convex lens as a magnifier, where must the object be placed?Read the answer and worked solution →A page is held 7.50 cm from a concave lens whose focal length is -10.0 cm. What image distance and magnification result?Read the answer and worked solution →What kind of image does a concave lens form, wherever the object is placed?Read the answer and worked solution →A camera lens of focal length 35.0 mm photographs a flower 75.0 cm away. How far from the lens must the film sit for a sharp image?Read the answer and worked solution →A slide projector has a lens of focal length 100 mm and the slide sits 103 mm from the lens. How far away is the screen that receives a sharp image?Read the answer and worked solution →A magnifying glass held 5.00 cm from a coin produces a magnification of 3.00. What is the focal length of the lens?Read the answer and worked solution →An object is held 25.0 cm from a lens of power -4.00 D, of the kind used to correct short sight. What magnification results?Read the answer and worked solution →Half of a lens is covered with black tape. What happens to the image the lens forms?Read the answer and worked solution →A small bulb filament is placed exactly at the focal point of a convex lens. What emerges from the other side?Read the answer and worked solution →Which set of statements describes the image formed by a flat mirror?Read the answer and worked solution →How is the focal length of a spherical mirror related to its radius of curvature?Read the answer and worked solution →What are the signs of the focal length and power for a concave mirror and for a convex mirror?Read the answer and worked solution →A concave mirror has a radius of curvature of 50.0 cm. What is its focal length?Read the answer and worked solution →The coils of a heater sit 27.3 cm from a concave mirror that forms their image 3.00 m in front of it. What is the magnification?Read the answer and worked solution →Shops use convex mirrors for security. What kind of image do they form and why does that help?Read the answer and worked solution →A makeup mirror produces a magnification of 1.50 when a face is 12.0 cm away. What is the focal length of the mirror?Read the answer and worked solution →A ray approaches a concave mirror parallel to its axis. Where does the reflected ray go?Read the answer and worked solution →A ray approaches a convex mirror parallel to its axis. How does the reflected ray behave?Read the answer and worked solution →Which pairing correctly matches mirrors with the lenses that behave like them?Read the answer and worked solution →A convex mirror has a radius of curvature of 3.00 cm. What are its focal length and its power?Read the answer and worked solution →A camera maker wants to replace an 800 mm focal length telephoto lens with a mirror. What radius of curvature must the mirror have?Read the answer and worked solution →A makeup mirror has a power of 1.50 D. What is its focal length?Read the answer and worked solution →Pull the rope to the right lensRead the answer and worked solution →Sort real and virtual imagesRead the answer and worked solution →Stack the optics termsRead the answer and worked solution →Match each optics termRead the answer and worked solution →Complete the optics rulesRead the answer and worked solution →

Worked practice questions

What is a virtual image?

  1. One that can be projected onto a screen
  2. One that is always inverted and smaller
  3. One that cannot be projected onto a screen
  4. One that forms only on a curved mirror
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Correct answer: C. One that cannot be projected onto a screen

1. A virtual image sits on the same side of the lens as the object and cannot be projected on a screen. 2. Light rays only appear to originate at a virtual image; they do not actually pass through that place in space. 3. A screen put there receives only diffuse light from the object rather than focused rays from the lens. 4. Virtual images can still be seen, because the lens of the eye converges the diverging rays into a real image on the retina. _Source: OpenStax College Physics (CC BY 4.0), Ch 25 "Geometric Optics", section 25.6 Image Formation by Lenses_

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In the magnification equation for a thin lens, what does a negative value of m mean?

  1. The image is upright relative to the object
  2. The image is inverted relative to the object
  3. The image lies behind the lens on the far side
  4. The image is smaller than the object itself
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Correct answer: B. The image is inverted relative to the object

1. Magnification is defined as m = image height divided by object height, and it also equals minus the image distance divided by the object distance. 2. Images that appear upright relative to the object have positive heights and those that are inverted have negative heights. 3. So a negative m means a negative image height, which means an inverted image. 4. The size of m, taken without its sign, says whether the image is larger or smaller than the object. 5. A magnification of -2.00 therefore means inverted and twice as tall, not upside down and half as tall. _Source: OpenStax College Physics (CC BY 4.0), Ch 25 "Geometric Optics", section 25.6 Image Formation by Lenses_

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A clear glass light bulb sits 0.750 m from a convex lens whose focal length is 0.500 m. What are the image distance and the magnification?

  1. 1.50 m and +2.00
  2. 0.300 m and -0.40
  3. -1.50 m and +2.00
  4. 1.50 m and -2.00
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Correct answer: D. 1.50 m and -2.00

1. Fix the sign convention first: the object distance do is positive, f is positive for this convex lens, and a negative image distance would mean a virtual image on the object's side. 2. The thin lens equation is 1/do + 1/di = 1/f, which rearranges to 1/di = 1/f - 1/do. 3. Substituting with units, 1/di = 1 divided by 0.500 m minus 1 divided by 0.750 m. 4. This is 2.000 per metre minus 1.333 per metre = 0.667 per metre. 5. Inverting, di = 1 divided by 0.667 per metre = 1.50 m, and the positive sign puts the image on the far side of the lens. 6. The magnification is m = minus di divided by do = minus 1.50 m divided by 0.750 m = -2.00. 7. The +2.00 option drops the minus sign built into the magnification equation and would wrongly call the image upright. 8. The -1.50 m option ignores that the object sits farther out than the focal length, which always gives a positive image distance. _Source: OpenStax College Physics (CC BY 4.0), Ch 25 "Geometric Optics", section 25.6 Image Formation by Lenses_

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