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Lesson 1: Refraction at a Boundary

Boundary Behavior

Refraction and Sight

The Cause of Refraction

Optical Density and Light Speed

The Direction of Bending

If I Were an Archer Fish


Lesson 2: The Mathematics of Refraction

The Angle of Refraction

Snell's Law

Ray Tracing and Problem-Solving

Determination of n Values


Lesson 3: Total Internal Reflection

Boundary Behavior Revisited

Total Internal Reflection

The Critical Angle


Lesson 4: Interesting Refraction Phenomena

Dispersion of Light by Prisms

Rainbow Formation

Mirages


Lesson 5: Image Formation by Lenses

The Anatomy of a Lens

Refraction by Lenses

Image Formation Revisited

Converging Lenses:

Ray Diagrams

Object-Image Relations

Diverging Lenses:

Ray Diagrams

Object-Image Relations

The Mathematics of Lenses


Lesson 6: The Eye

The Anatomy of the Eye

Image Formation and Detection

The Wonder of Accommodation

Farsightedness and its Correction

Nearsightedness and its Correction


Lesson 5: Image Formation by Lenses

The Mathematics of Lenses

Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a lens. The use of these diagrams were demonstrated earlier in Lesson 5 for both converging and diverging lenses. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. To obtain this type of numerical information, it is necessary to use the Lens Equation and the Magnification Equation. The lens equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:

The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:

These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.

 

As a demonstration of the effectiveness of the lens equation and magnification equation, consider the following sample problem and its solution.

Sample Problem #1

A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size.

Like all problems in physics, begin by the identification of the unknown information.

ho = 4.00 cm
do = 45.7 cm
f = 15.2 cm

Next identify the unknown quantities which you wish to solve for.

di = ???
hi = ???

To determine the image distance, the lens equation must be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

1/f = 1/do + 1/di

1/(15.2 cm) = 1/(45.7 cm) + 1/di

0.0658 cm-1 = 0.0219 cm-1 + 1/di

0.0439 cm-1 = 1/di

di = 22.8 cm

The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

hi/ho = - di/do

hi /(4.00 cm) = - (22.8 cm)/(45.7 cm)

hi = - (4.00 cm) • (22.8 cm)/(45.7 cm)

hi = -1.99 cm

The negative values for image height indicate that the image is an inverted image. As is often the case in physics, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image height, a negative value always indicates an inverted image.

From the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 45.7 cm from a double convex lens having a focal length of 15.2 cm, then the image will be inverted, 1.99-cm tall and located 22.8 cm from the lens. The results of this calculation agree with the principles discussed earlier in this lesson. In this case, the object is located beyond the 2F point (which would be two focal lengths from the lens) and the image is located between the 2F point and the focal point. This falls into the category of Case 1 : The object is located beyond 2F for a converging lens.

 

Now lets try a second sample problem:

Sample Problem #2

A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length of 15.2 cm. (NOTE: this is the same object and the same lens, only this time the object is placed closer to the lens.) Determine the image distance and the image size.

 

Again, begin by the identification of the unknown information.

ho = 4.00 cm
do = 8.3 cm
f = 15.2 cm

Next identify the unknown quantities which you wish to solve for.

di = ???
hi = ???

To determine the image distance, the lens equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

1/f = 1/do + 1/di

1/(15.2 cm) = 1/(8.30 cm) + 1/di

0.0658 cm-1 = 0.120 cm-1 + 1/di

-0.0547 cm-1 = 1/di

di = -18.3 cm

The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

hi/ho = - di/do

hi /(4.00 cm) = - (-18.3 cm)/(8.30 cm)

hi = - (4.00 cm) • (-18.3 cm)/(8.30 cm)

hi = 8.81 cm

The negative value for image distance indicates that the image is a virtual image located on the object's side of the lens. Again, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always means the image is located on the object's side of the lens. Note also that the image height is a positive value, meaning an upright image. Any image which is upright and located on the object's side of the lens is considered to be a virtual image.

From the calculations in the second sample problem it can be concluded that if a 4.00-cm tall object is placed 8.30 cm from a double convex lens having a focal length of 15.2 cm, then the image will be enlarged, upright, 8.81-cm tall and located 18.3 cm from the lens on the object's side. The results of this calculation agree with the principles discussed earlier in this lesson. In this case, the object is located in front of the focal point (i.e., the object distance is less than the focal length) and the image is located behind the lens. This falls into the category of Case 5: The object is located in front of F (for a converging lens).

 

The third sample problem will pertain to a diverging lens.

 

Sample Problem #3

A 4.00-cm tall light bulb is placed a distance of 35.5 cm from a diverging lens having a focal length of -12.2 cm. Determine the image distance and the image size.

Like all problems in physics, begin by the identification of the unknown information.

ho = 4.00 cm
do = 35.5 cm
f = -12.2 cm
 

Next identify the unknown quantities which you wish to solve for.

di = ???
hi = ???

To determine the image distance, the lens equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

1/f = 1/do + 1/di

1/(-12.2 cm) = 1/(35.5 cm) + 1/di

-0.0820 cm-1 = 0.0282 cm-1 + 1/di

-0.110 cm-1 = 1/di

di = -9.08 cm

The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

hi/ho = - di/do

hi /(4.00 cm) = - (-9.08 cm)/(35.5 cm)

hi = - (4.00 cm) * (-9.08 cm)/(35.5 cm)

hi = 1.02 cm

The negative values for image distance indicates that the image is located on the object's side of the lens. As mentioned, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always indicates the existence of a real image located on the object's side of the lens. In the case of the image height, a positive value indicates an upright image.

From the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 35.5 cm from a diverging lens having a focal length of 12.2 cm, then the image will be upright, 1.02-cm tall and located 9.08 cm from the lens on the object's side. The results of this calculation agree with the principles discussed earlier in this lesson. Diverging lenses always produce images which are upright, virtual, reduced in size, and located on the object's side of the lens.

 

 

Sign Conventions

The sign conventions for the given quantities in the lens equation and magnification equations are as follows:

  • f is + if the lens is a double convex lens (converging lens)
  • f is - if the lens is a double concave lens (diverging lens)
  • di is + if the image is a real image and located on the opposite side of the lens.
  • di is - if the image is a virtual image and located on the object's side of the lens.
  • hi is + if the image is an upright image (and therefore, also virtual)
  • hi is - if the image an inverted image (and therefore, also real)

 

Like many mathematical problems in physics, the skill is only acquired through much personal practice. Perhaps you would like to take some time to try the following problems.

 

 

Check Your Understanding

1. Determine the image distance and image height for a 5-cm tall object placed 45.0 cm from a double convex lens having a focal length of 15.0 cm.

 

 

2. Determine the image distance and image height for a 5-cm tall object placed 30.0 cm from a double convex lens having a focal length of 15.0 cm.

 

 

3. Determine the image distance and image height for a 5-cm tall object placed 20.0 cm from a double convex lens having a focal length of 15.0 cm.

 

 

4. Determine the image distance and image height for a 5-cm tall object placed 10.0 cm from a double convex lens having a focal length of 15.0 cm.

 

 

5. A magnified, inverted image is located a distance of 32.0 cm from a double convex lens with a focal length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual.

 

 

6. ZINGER: An inverted image is magnified by 2 when the object is placed 22 cm in front of a double convex lens. Determine the image distance and the focal length of the lens.

 

 

7. A double concave lens has a focal length of -10.8 cm. An object is placed 32.7 cm from the lens's surface. Determine the image distance.

 

 

8. Determine the focal length of a double concave lens which produces an image which is 16.0 cm behind the lens when the object is 28.5 cm from the lens.

 

 

9.A 2.8-cm diameter coin is placed a distance of 25.0 cm from a double concave lens which has a focal length of -12.0 cm. Determine the image distance and the diameter of the image.

 

 

10. The focal point is located 20.0 cm from a double concave lens. An object is placed 12 cm from the lens. Determine the image distance.

 

 

 

 

 

Lesson 5: Image Formation by Lenses

 

Go to Lesson 6
 

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