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Chapter 23: Problem 84

An object of height \(5.00 \mathrm{cm}\) is placed \(20.0 \mathrm{cm}\) from a converging lens of focal length \(15.0 \mathrm{cm} .\) Draw a ray diagram to find the height and position of the image.

Short Answer

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Question: An object of height 5.0 cm is placed 20.0 cm from a converging lens with a focal length of 15.0 cm. Determine the position and the height of the image formed and provide steps to draw the ray diagram. Answer: The position of the image is 12.0 cm on the opposite side of the lens and it has a height of 3.0 cm (upright). To draw the ray diagram, follow these steps: 1. Draw the lens, marking its focal points F1 and F2, and the principal axis (PA) passing through the center of the lens. 2. Mark the position of the object (20.0 cm from the lens and 5.0 cm above the principal axis). 3. Draw a ray from the top of the object to the lens parallel to the principal axis. After passing through the lens, the ray should go through the focal point F2. 4. Draw another ray from the top of the object going through the center of the lens. This ray will continue without bending on the other side of the lens. 5. Mark the intersection of the two rays on the opposite side of the lens, which will be the position of the image (12.0 cm from the lens and 3.0 cm above the principal axis, upright).

Step by step solution

01

Lens Formula

The lens formula for a converging lens is given by: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\) Where: f is the focal length of the lens (15.0 cm) u is the object distance (-20.0 cm, negative since the object is on the same side as the incoming light) v is the image distance, which needs to be found First, plug the known values into the lens formula: \(\frac{1}{15} = \frac{1}{-20} + \frac{1}{v}\)
02

Solve for the Image Distance (v)

To solve for v, first simplify the equation by finding the common denominator, which is 60: \(\frac{20}{60}=\frac{-15}{60}+\frac{1}{v}\) Now, combine the fractions on the left side: \(\frac{5}{60}=\frac{1}{v}\) Find the inverse of both sides to determine the value of v: \(v=\frac{60}{5}\) \(v=12\mathrm{cm}\) (positive, which indicates the image is formed on the opposite side of the lens)
03

Magnification Formula

The magnification formula is given by: \(m=\frac{h'}{h}=\frac{-v}{u}\) Where: h is the height of the object (5.0 cm) h' is the height of the image (which needs to be found) u is the object distance (-20.0 cm) v is the image distance (12.0 cm) First, plug the known values into the formula: \(m=\frac{-v}{u}=\frac{-12}{-20}\)
04

Solve for the Height of the Image (h')

Simplify the magnification: \(m=0.6\) Now, use the magnification to find the height of the image (h'): \(h'=mh\) \(h'=0.6\times5.0\mathrm{cm}\) \(h'=3.0\mathrm{cm}\) (positive, which means the image is upright) Now, we have the position of the image (12.0 cm on the opposite side of the lens) and the height of the image (3.0 cm, upright). To draw the ray diagram, follow these steps: 1. Draw the lens, marking its focal points F1 and F2, and the principal axis (PA) passing through the center of the lens. 2. Mark the position of the object (20.0 cm from the lens and 5.0 cm above the principal axis). 3. Draw a ray from the top of the object to the lens parallel to the principal axis. After passing through the lens, the ray should go through the focal point F2. 4. Draw another ray from the top of the object going through the center of the lens. This ray will continue without bending on the other side of the lens. 5. Mark the intersection of the two rays on the opposite side of the lens, which will be the position of the image (12.0 cm from the lens and 3.0 cm above the principal axis, upright).

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Most popular questions from this chapter

An object that is \(6.00 \mathrm{cm}\) tall is placed \(40.0 \mathrm{cm}\) in front of a diverging lens. The magnitude of the focal length of the lens is \(20.0 \mathrm{cm} .\) Find the image position and size. Is the image real or virtual? Upright or inverted?
The focal length of a thin lens is \(-20.0 \mathrm{cm} .\) A screen is placed \(160 \mathrm{cm}\) from the lens. What is the \(y\) -coordinate of the point where the light ray shown hits the screen? The incident ray is parallel to the central axis and is \(1.0 \mathrm{cm}\) from that axis.
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 \(\sin \theta=\tan \theta \approx \theta(\text { in radians) } .]\)
Apply Huygens's principle to a 5 -cm-long planar wavefront approaching a reflecting wall at normal incidence. The wavelength is \(1 \mathrm{cm}\) and the wall has a wide opening (width \(=4 \mathrm{cm}\) ). The center of the incoming wavefront approaches the center of the opening. Repeat the procedure until you have wavefronts on both sides of the wall. Without worrying about the details of edge effects, what are the general shapes of the wavefronts on each side of the reflecting wall? Repeat Problem 2 for an opening of width $0.5 \mathrm{cm} .$
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