Questions: 1 2 3 4 5 6 7

Images formed by Lenses Solutions

1. Use the Principle Ray Technique to find the image, created by a converging lens, of an object placed between the first focal point F and the lens. Characterize the image.
   
   

   The image, I, is erect, larger, and virtual. It is located between the left focal point and the object at O. We assume that the observer in on the right side.
   

   

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2. Use the Principle Ray Technique to find the image, created by a diverging lens, of an object placed between F and the lens. Characterize the image.
   
   

   The image, I, is erect, smaller, and virtual. It is located between the object, O, and the lens.
   

   

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3. The convex lens of a magnifying glass has a focal length of 20 cm. At what distance from the postage stamp must you hold this lens if the image of the stamp is to be twice as large as the stamp and
   (a) the image is inverted, or
   (b) the image is erect?

   For a convex lens, f = +20 cm.
   

   (a) Given that the image is inverted and twice as large, we have M = -2. Now our definition of magnification is M = -i/o, therefore i = 2o. Using the lens formula

   1/o + 1/i = 1/f .

   Substituting i = 2o , this becomes
   

   1/o + 1/2o = 1/f ,

   or
   

   o = 3f/2 = 3(20 cm)/2 = 30 cm .

   Thus i = 2(30 cm) = 60 cm.

   (b) Since the image is erect and twice as large, then M = +2. Therefore i = -2o. Substituting this into the lens equation

   1/o - 1/2o = 1/f ,

   yields

   o = ½f = ½(20 cm) = 10 cm .

   Thus i = -2(10 cm) = -20 cm.
   

   

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4. A lens forms an image of an object. The object is 20.0 cm from the lens. The image is 5.00 cm from the lens on the same side as the object.
   (a) What is the focal length of the lens? Is the lens converging or diverging?
   (b) If the object is 2.00 cm tall, how tall is the image? Is it erect or inverted?
   

   (a) We given that o = 20.0 cm and that i = -5.00 cm. The minus sign for i indicates that the image is on the same side of the lens as the object. Using the lens equation

   1/f = 1/o + 1/i = 1/20 + 1/(-5) = -3/20 .

   The focal length of the lens is thus -20/3 = -6.67 cm. The minus sign indicates that the lens is diverging or concave.
   

   The magnification is given by

   M = -i/o = -(-5)/20 = ¼ .

   As well, the size of the image is given by
   

   H_image = Mh_object = ¼(2.00 cm) = 0.50 cm .

   The magnification is positive so the image is erect. As well the image is on the same side as the object, so it is virtual and virtual images are erect.

   

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5. An optical system comprises an object, of height 3 mm, fixed at a distance of 1.60 m from a screen. A converging lens can be moved freely in the space between the object and the screen. It is found that there are TWO positions of the lens which will give a real image on the screen, and that these two positions are 80 cm apart. Determine the focal length of the lens and the sizes of the two images. A neat diagram is required. Think about the symmetry of the problem.
   

   We are dealing with two cases, so we need two diagrams and one equation for each case.
   

   

   Our equations for both cases are
   

   1/o₁ + 1/i₁ = 1/f , (1)

   and
   

   1/o₂ + 1/i₂ = 1/f . (2)

   The other information that we are given, and shown in the diagram, is
   

   |o₁| + |i₁| = D , (3)
   |o₂| + |i₂| = D , (4)

   and
   

   |o₂| - |o₁| = L . (5)

   The absolute signs on the object and image distances are there to cancel the sign convention. They are not absolutely necessary here since the images are on the transmission side and thus must be positive. The objects are real objects and are also already positive.
   

   We next write i₁, i₂, and o₂ in terms of D, L, and o₁,
   

   i₁ = D - o₁ ,
   o₂ = L + o₁ ,

   and
   

   i₂ = L - o₁ .

   We substitute these into equations (1) and (2),
   

   1/o₁ + 1/(D - o₁) = 1/f , (1a)

   and
   

   1/(L + o₁) + 1/( L - o₁) = 1/f . (2a)

   Equating (1a) and (1b) yields,
   

   1/o₁ + 1/(D - o₁) = 1/(L + o₁) + 1/(L - o₁) .

   Rationalizing each side yields
   

   D / o₁(D - o₁) = 2L / (L + o₁)(L - o₁) .

   Cross-multiplying gives
   

   D(L + o₁)(L - o₁) = 2Lo₁(D - o₁) .

   Rearranging yields
   

   (D-2L)o_i² - 2LDo₁ + DL² = 0 .

   This quadratic simplifies further since we are given that L = ½D. We find
   

   o₁ = ¼D = +40 cm .

   Our other values are i₁ = +120 cm, o₂ = +120 cm, and i₂ = +40 cm.

   The magnification of the image in the first case is
   

   M₁ = - i₁/o₁ = - 120/40 = -3 .

   The first image is 3 times larger, inverted, and real.
   

   The magnification of the image in the second case is
   

   M₂ = - i₂/o₂ = - 40/120 = -1/3 .

   The image is one-third of the original size, inverted, and real.
   

   Notice that the image and object distances in the second case are the reverse of the first. This make sense because of the symmetry of light rays. We find an image by following rays from the object to where they intercept at the image. We could imagine that the image was an object and follow the rays back to the object. The second case is equivalent to swapping the image and object.
   

   

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6. Two converging lenses have focal lengths f₁ = 30 cm and f₂ = 20 cm. They are placed 60 cm apart along the same axis and an object is placed 50 cm from lens #1 on the side opposite lens #2. Locate the final image and calculate its overall magnification. State the final image characteristics relative to the original object.
   

   We start with the first lens. We are given that o₁ = 50 cm and f₁ = 30 cm. We find i₁ from
   

   1/o₁ + 1/i₁ = 1/f₁ .

   We find
   

   i₁ = f₁o₁ / (o₁ - f₁) = (30)(50)/(50 - 30) = 75 cm .

   The image formed by the first lens is the object for the second lens. Since the lenses are 60 cm apart, the "object" is 15 cm away from lens #2 on the transmission side. The "object" is thus virtual and
   

   o₂ = -15 cm .

   We are given that f₂ = 20 cm. We find i₂ from
   

   1/o₂ + 1/i₂ = 1/f₂ .

   We find
   

   i₂ = f₂o₂ / (o₂ - f₂) = (20)(-15)/(-15 - 20) = +8.57 cm .

   The final image is on the transmission side of the second lens and is thus real.
   

   The overall magnification is given by
   

   M = M₁ M₂ = (-i₁/o₁) (-i₁/o₁) = -(+75)/(+50) -(-15)/(+8.57) = -0.86 .

   The final image is inverted and 86% of the original size.
   

   Drawing the path of the rays is a tricky business as shown below. First we draw the Principle Ray, 1, 2, and, 3 in cyan to find the image formed by the first lens. Note that the lines are dashed as they pass through the second lens; they never reach there but are refracted by the second lens. Ray 1 is parallel to the axis, so it will be refracted through the second lens so that it passes through the second lens' transmission side focal point. This is shown as red ray 1.
   

   The principle rays are not the only rays emitted by the real object. One ray, 4, will pass through the focal point of the second lens on its way to I₁, is refracted by lens #2, and come out parallel to the axis. This is shown as red ray 4.
   

   Another ray, 5, will pass through the centre of the second lens on its way to I₁ and not be deviated at all. This is shown as red ray 5.
   

   

   

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7. For the optical system shown below, determine the final image position as seen by the observer of the object at position O, due to the lenses and the plane mirror.
   

   

   We start with the first lens. We are given that o₁ = 15 cm and f₁ = 10 cm. We find i₁ from
   

   1/o₁ + 1/i₁ = 1/f₁ .

   We find
   

   i₁ = f₁o₁ / (o₁ - f₁) = (10)(15)/(15 - 10) = 30 cm .

   The image formed by the first lens is the object for the second lens. Since the lenses are 40 cm apart, the "object" is 10 cm away from lens #2 on the incident side. The "object" is thus real and
   

   o₂ = +10 cm .

   We are given that f₂ = 5 cm. We find i₂ from
   

   1/o₂ + 1/i₂ = 1/f₂ .

   We find
   

   i₂ = f₂o₂ / (o₂ - f₂) = (5)(10)/(10 - 5) = +10 cm .

   The image formed by the second lens is the object for the plane mirror. The "object" is 30 cm away from the mirror, so an observer looking at the mirror will see a reflected image 30 cm behind the mirror.
   

   The final image is behind of the mirror so it is virtual.
   

   The overall magnification is given by
   

   M	= M1 × M2 × Mmirror
   	= (-i1/o1) × (-i1/o1) × 1
   	= -(+30)/(+15) × -(+10)/(+10) × 1
   	= +2

   The final image has the same orientation as the object and is twice as big.
   

   Note how crucial it is in this problem to know exactly where the observer is.
   

   What would you have to do if the observer was on the same side of lens #1 as the object and was looking through the lens?

   

Questions? mike.coombes@kpu.ca