
Lenses use curved surfaces to focus light rays into an image. The focal length of a lens depends on the curvature of the surface of the lens and the index of refraction of the lens material. Converging lenses can create real or virtual images, depending on the object distance relative to the focal length of the lens. Real images occur when light rays physically meet at a point; virtual images occur when the light rays only appear to meet at a point. The thin lens formula is used to calculate object distance, image distance, or focal length when the other two quantities are known.

focal length, focal point, convex lens, concave lens, real image, virtual image


$$\frac{1}{{d}_{o}}+\frac{1}{{d}_{i}}=\frac{1}{f}$$

 $$m=\frac{{d}_{i}}{{d}_{o}}$$



Review problems and questions 

 Does a flat mirror produce real or virtual images?

A flat mirror can only produce a virtual image. The image is behind the mirror where no real light rays can reach, and where you cannot project it onto a piece of paper.

 Does a concave lens produce real images, virtual images, or both?

A concave (or diverging) lens can only produce virtual images.

 A thin convex lens produces an image that is magnified and inverted. Where is the object located?

The object is located between one and two focal lengths from the lens.

 Cassandra has a lens with a focal length of 20 cm. Where should she place an object to get an image at a distance of 60 cm?

Answer: She should place it at an object distance of 30 cm.
Asked: object distance d_{o} Given: focal length f = 20 cm; image distance d_{i} = 60 cm Relationships: $$\frac{1}{{d}_{o}}+\frac{1}{{d}_{i}}=\frac{1}{f}$$Solve:$$\begin{array}{l}\frac{1}{{d}_{o}}+\frac{1}{60\text{cm}}=\frac{1}{20\text{cm}}\\ \frac{1}{{d}_{o}}=\frac{1}{30\text{cm}}\\ {d}_{o}=30\text{cm}\end{array}$$

 Damien places an object at a distance of 90 cm from a lens, and the image distance is 30 cm.
 What is the focal length of his lens?
 What is the magnification?

Answer:  The focal length is 22.5 cm.
 The magnification is −1/3.
Solution:
 Asked: focal length f
Given: object distance d_{o} = 90 cm; image distance d_{i} = 30 cm Relationships: $$\frac{1}{{d}_{o}}+\frac{1}{{d}_{i}}=\frac{1}{f}$$ Solve: $$\begin{array}{l}\frac{1}{90\text{cm}}+\frac{1}{30\text{cm}}=\frac{1}{f}\\ \frac{1}{f}=\frac{4}{\text{90cm}}\\ f=22.5\text{cm}\end{array}$$  Asked: magnification m
Given: object distance d_{o} = 90 cm; image distance d_{i} = 30 cm Relationships: $$m=\frac{{d}_{i}}{{d}_{o}}$$Solve: $$\begin{array}{l}m=\frac{30\text{cm}}{90\text{cm}}=\frac{1}{3}\end{array}$$

 A onedollar bill is 6.5 cm high by 15.5 cm wide. When the bill is placed 26 cm away from a thin convex lens, an inverted image of the dollar bill is projected onto the wall 1.0 m away from the bill on the far side of the lens.
 Is the image real or virtual? How do you know?
 What is the image distance d_{i} ?
 What are the dimensions of the image?

Answer:  The image is real. Only real images can be projected onto a surface.
 The image distance d_{i} is 0.74 cm.
 The image is 19 cm high and 44 cm wide.

 A simple projector shines light through a slide to project an enlarged image of the slide on the wall. Slides with a width of 6 cm are placed 11 cm, behind a lens with a focal length of 10 cm.
 Where does the image appear? How far should the projector be from the wall?
 What is the magnification?
 How large is the image?
 What does the orientation of the image mean?

Answer:  The image is projected on the wall 110 cm away.
 The magnification is −10.
 The image is 60 cm wide and inverted.
 Because the image is inverted, the slide should be inserted upsidedown into the projector.
Solution:  Asked: image distance d_{i}
Given: distance of the slide, which is the object distance d_{o} = 11 cm; focal length f = 10 cm Relationships: $$\frac{1}{{d}_{o}}+\frac{1}{{d}_{i}}=\frac{1}{f}$$ Solve: Rearrange and solve the equation for 1⁄d_{i}:$$\begin{array}{c}\frac{1}{{d}_{i}}=\frac{1}{f}\frac{1}{{d}_{o}}=\frac{1}{10\text{cm}}\frac{1}{11\text{cm}}=\frac{1}{110\text{cm}}\\ \frac{1}{{d}_{i}}=\frac{1}{110\text{cm}}\end{array}$$Take the inverse of both sides to solve for d_{i}:$${d}_{i}={\left(\frac{1}{110\text{cm}}\right)}^{1}=110\text{cm}$$ Answer: The image is projected on the wall 110 cm away.  Asked: magnification m of this system
Given: object distance d_{o} = 11 cm; image distance d_{i} = 110 cm (previous answer) Relationships: $$m=\frac{{d}_{i}}{{d}_{o}}$$Solve:$$m=\frac{{d}_{i}}{{d}_{o}}=\frac{110\text{cm}}{11\text{cm}}=10$$Answer: The magnification is −10.  Asked: image size h_{i}
Given: object size (slide) h_{o} = 6 cm; magnification m = −10 Relationships: $$m=\frac{{d}_{i}}{{d}_{o}}$$Solve: Rearrange and solve equation for h_{i}:$${h}_{i}=m\times {h}_{o}=\text{}10\times 6\text{cm}=60\text{cm}$$Answer: The image is 60 cm wide and inverted.  Because the image is inverted, the slide should be inserted upsidedown into the projector.

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