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Chapter 23: Problem 66

Sketch a ray diagram to show that if an object is placed less than the focal length from a converging lens, the image is virtual and upright. ( Will tutorial: magnifying glass)

Short Answer

Expert verified
Answer: When an object is placed closer to a converging lens than its focal length, the image formed is virtual and upright. Additionally, the virtual image is larger compared to the object and is located on the same side as the object.

Step by step solution

01

Identify the lens type and focal point

A converging lens is thicker at the middle than at the edges, which causes the light rays to bend toward each other and converge at a point. This point is called the focal point. It is important to notice that a converging lens has two focal points, one on each side of the lens. However, in this problem, we will be considering only the principal focal point that lies on the side of the light rays coming out of the lens.
02

Draw the lens and principal axis

Draw a vertical line representing the converging lens and draw a horizontal line passing through the center of the lens. This horizontal line is called the principal axis.
03

Place the object closer to the lens than its focal length

Place an arrow vertically above the principal axis on the left side of the lens. This arrow represents the object. Make sure to position the object closer to the lens than its focal length, and label this distance as 'd_o', where 'd_o' is less than the focal length 'f'.
04

Draw the three principal rays

From the tip of the object, draw the following three principal rays: 1. Ray 1: Passes through the center of the lens without bending, and continues straight on the other side. 2. Ray 2: Parallel to the principal axis until it reaches the lens, then refracts and passes through the principal focal point on the other side of the lens. 3. Ray 3: Passes through the principal focal point on the object's side of the lens, then refracts to be parallel to the principal axis on the other side of the lens.
05

Notice that the rays do not converge on the right side of the lens

As you can see, the three principal rays diverge after passing through the lens. This means that the image is not formed on the right side of the lens. Instead, we need to trace these rays back behind the lens to find the virtual image.
06

Trace the rays to form the virtual image

Extend the Ray 1 and Ray 2 back behind the lens until they intersect at a point. This point of intersection will be the location of the virtual image. Draw an arrow vertically above the principal axis at this point to represent the virtual image.
07

Observe the image properties

The virtual image formed is located on the same side as the object, and it is larger and upright compared to the object. This confirms that when an object is placed closer to a converging lens than its focal length, the image formed is virtual and upright. You have now successfully sketched a ray diagram to show that when an object is placed closer to a converging lens than its focal length, the resulting image is virtual and upright.

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Most popular questions from this chapter

A glass lens has a scratch-resistant plastic coating on it. The speed of light in the glass is \(0.67 c\) and the speed of light in the coating is \(0.80 c .\) A ray of light in the coating is incident on the plastic-glass boundary at an angle of \(12.0^{\circ}\) with respect to the normal. At what angle with respect to the normal is the ray transmitted into the glass?
A beam of light in air is incident on a stack of four flat transparent materials with indices of refraction 1.20 \(1.40,1.32,\) and \(1.28 .\) If the angle of incidence for the beam on the first of the four materials is \(60.0^{\circ},\) what angle does the beam make with the normal when it emerges into the air after passing through the entire stack?
Show that the deviation angle \(\delta\) for a ray striking a thin converging lens at a distance \(d\) from the principal axis is given by \(\delta=d l f .\) Therefore, a ray is bent through an angle \(\delta\) that is proportional to \(d\) and does not depend on the angle of the incident ray (as long as it is paraxial). [Hint: Look at the figure and use the small-angle approximation \(\sin \theta=\tan \theta \approx \theta(\text { in radians) } .]\)
A concave mirror has a radius of curvature of \(5.0 \mathrm{m} .\) An object, initially \(2.0 \mathrm{m}\) in front of the mirror, is moved back until it is \(6.0 \mathrm{m}\) from the mirror. Describe how the image location changes.
A \(45^{\circ}\) prism has an index of refraction of \(1.6 .\) Light is normally incident on the left side of the prism. Does light exit the back of the prism (for example, at point \(P\) )? If so, what is the angle of refraction with respect to the normal at point \(P ?\) If not, what happens to the light?
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