Optics and Ray Diagrams for Exam Review

Blaming and Problem Solving

• The speaker jokingly suggests blaming the teacher for failures on tests.
• The speaker refers to himself as a "blame doctor."

Speed Calculation

• The speaker presents a problem involving calculating speed, using the formula $c = 3 \times 10^8$ (where $c$ is the speed of light).
• The speaker explains how to cross-multiply to solve for an unknown variable $b$:
  $1.27 \times b = 3 \times 10^8$
  $b = \frac{3 \times 10^8}{1.27}$
• The speaker guides the student through the calculation, emphasizing the importance of using the calculator correctly.
• The result of the division $3 / 1.27$ is approximately $2.36$, so $b = 2.36 \times 10^8$
• The units for speed are implied to be meters per second.

PhET Lab Simulation

• The speaker transitions to the topic of optics, specifically convex lenses.
• The speaker encourages the use of PhET Lab simulations for visualization but discourages using Google during quizzes, as screen monitoring will occur to prevent cheating.
• The PhET Lab activity is named "Lenses and Mirrors."

Ray Diagrams and Convex Lenses

• The speaker discusses how to construct ray diagrams for lenses.
• Given: Focal length $f = 5$ cm and object distance $d_o = 10$ cm.
• The formula for image distance $d<em>i$ is introduced. The reciprocal of the image distance equals the reciprocal of the focal length, minus the reciprocal of the object distance. $\frac{1}{d</em>i} = \frac{1}{f} - \frac{1}{d_o}$
• Which can be rewritten as: $d<em>i = \frac{1}{\frac{1}{f} - \frac{1}{d</em>o}}$
• When you solve for the inverse, the formulas becomes: $\frac{1}{d<em>i} = \frac{1}{5} - \frac{1}{10}$ giving $d</em>i = 10$ cm.
  • This means with an object at 10cm and a focal point at 5cm, the image distance is at 10cm.
• A step-by-step process for drawing a ray diagram is provided: draw a line from the object to the lens, then from the lens through the focal point. The image will appear where the lines intersect.
• In this specific case (object at 10 cm, focal length of 5 cm), the image is: real, inverted, and the same size as the object.

Concave Lenses

• The speaker moves on to concave lenses.
• With a concave lens, the image is always upright and virtual.
• Ray Diagram for Concave Lens: The focal point is located. Light rays diverge through the focal point. Extend the rays backward to find the image location.

Convex Lens Image Characteristics

• If an object is further away from a convex lens than the focal point, the image is upside down and real.
• If an object is closer to a convex lens than the focal point, the image is upright and virtual, located behind you.
• If the object is at the focal point, there is no image.

Converging Lens Problem

• A converging lens has a focal length of $f = 0.45$ meters.
• An object is placed at $d_o = 0.11$ meters.
• The image distance $d<em>i$ is calculated using the lens equation: $\frac{1}{d</em>i} = \frac{1}{f} - \frac{1}{d<em>o}$$d</em>i = \frac{1}{\frac{1}{0.45} - \frac{1}{0.11}}$
• The image distance is negative: $d_i = -0.15$ meters.
• A negative $d_i$ indicates a virtual image.

More Calculations

• Object height: $h_o = 0.33$ meters.

• Object distance: $d_o = 0.40$ meters from the converging lens.

• Focal length: $f = 0.12$ meters.

• Calculating $d<em>i$: $\frac{1}{d</em>i} = \frac{1}{f} - \frac{1}{d<em>o}$$\frac{1}{d</em>i} = \frac{1}{0.12} - \frac{1}{0.40}$
  $d_i = 0.17$

• The calculated $d_i$ is positive (0.17), indicating a real image.

• Using the magnification formula:
  $\frac{h<em>i}{h</em>o} = -\frac{d<em>i}{d</em>o}$

• Solve the height of the image: $h<em>i \times d</em>0 = -d<em>i \times h</em>o$
  $h<em>i = \frac{-d</em>i \times h<em>o}{d</em>0}$

• Substitution:
  $h<em>i = \frac{-0.17 \times 0.33}{0.40}$$h</em>i = -0.14$

• Since $d_i$ is positive and height is negative, the image is real and inverted.

Positive Attitude

• A personal anecdote about helping a student who initially disliked physics to appreciate it.