Two lenses that are 35 cm apart are used to form an image,...

Two lenses that are 35 cm apart are used to form an image, as shown in Figure 17.48. Lens 1 is converging and has the focal length f1 = 14 cm; lens 2 is diverging and has the focal length f2 = -7.0 cm. An object is placed 24 cm to the left of lens 1. (a) Use a ray diagram to find the approximate location of the image produced by the two lenses. (The image of lens 1 is the object for lens 2.) (b) Is the final image upright or inverted? (c) Is the final image real or virtual? Explain.

Two lenses that are 35 cm apart are used to form an
image, as shown in Figure 17.48. Lens 1 is converging
and has the focal length f1 = 14 cm; lens 2 is diverging
and has the focal length f2 = -7.0 cm. An object is
placed 24 cm to the left of lens 1. (a) Use a ray diagram
to find the approximate location of the image produced
by the two lenses. (The image of lens 1 is the object for
lens 2.) (b) Is the final image upright or inverted? (c) Is
the final image real or virtual? Explain.

Key Concepts

Thin Lens Equation

This is the fundamental relation 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the object distance from the lens, and di is the image distance. It is critical in determining where an image will form relative to a given lens and is the basis for analyzing lens systems.

Compound Lens System

In a multi-lens system, the image produced by the first lens acts as the object for the subsequent lens. Understanding this sequential process is essential for determining the final image position, orientation, and nature, whether real or virtual.

Ray Diagrams

Ray diagrams are schematic representations that trace the path of light through optical systems. They help visualize how rays converge or diverge after interacting with lenses, and they are useful for qualitatively predicting the position, orientation, and type (real or virtual) of the resultant image.

Real vs Virtual Images

This concept differentiates images that are formed by actual crossing of light rays (real images) from those that appear to be located where light rays seem to diverge from (virtual images). This distinction is key in optics, especially in the interpretation of ray diagrams and in the analysis of lens systems.

Image Orientation and Magnification

Magnification is used to determine the size change and orientation (upright or inverted) of the image relative to the object. The sign and magnitude of the magnification indicate whether the image is inverted or upright and whether it is enlarged or reduced, which is particularly important when analyzing compound optical setups.

Transcript

00:01 Now, given in the question, two lenses are separated by 35 centimeters.

00:12 Lens 1 is converging having a focal length of plus 14 centimeters.

00:21 Lens 2 is diverging having a focal length of minus 7 centimeters.

00:33 An object is placed 24 centimeter to the left of lens 1, which means the object distance for lens 1, let it be u1, is 24 centimeters.

00:55 It has been asked to find, first, a, use a ray diagram to find the approximate location of the image produced by the two lenses.

01:08 So, it has been asked to estimate the approximate location of the image produced by the two lenses using ray diagram.

01:21 Second, it has been asked to ascertain whether the final image is upright or inverted.

01:31 And third, whether the final image is real or virtual.

01:42 So let this be the common principle axis to both the lenses.

01:48 And this is our first lens, which is converging, the lens 1, which is converging.

01:54 The focal length, focal focus point is f1 on both sides of the lenses.

02:00 And the focal length is 14 centimeters.

02:03 F1 is 14 centimeters.

02:07 Let us say an object is placed, an object is placed at a distance 24 centimeter to the left of this first lens.

02:19 Now, obviously, a ray parallel to the principle axis will pass through the focus and a ray passing through the optical center will pass undeviated.

02:29 They meet at a certain point that would be the position of the image corresponding to the first lens.

02:39 Now we can use lens equation to accurately position, to accurately find the position of the image formed due to the first lens...

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