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AP Physics 2 : Thin Lens Equation

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Example Questions

Example Question #1 : Optics

Suppose a student who was farsighted wears glasses that allows him to read at a distance of 20cm from his eyes to the book. His near-point distance is 63cm . If his glasses are 1.5cm from his eyes, what is the refractive power of his glasses lenses?

Possible Answers:

3.846D

6.346D

1.613D

4.536D

2.342D

Correct answer:

3.846D

Explanation:

Write the thin-lens equation.

$\frac{1}{f}= \frac{1}{d_o}+\frac{1}{d_i}$

d_o represents the object distance from the lens, subtracting the distance between the student's eyes and the lens.

d_o=20cm-1.5cm=18.5cm

The image distance is:

d_i=1.5cm-63cm=-61.5cm

Substitute the givens to the equation.

$\frac{1}{f}= \frac{1}{18.5}+\frac{1}{-61.5}= 0.037794 cm^{-1}$

$f=\frac{1}{0.037794cm}=26.4592cm = 0.26 m$

Refractive power is the inverse of focal length in .

$\frac{1}{0.26m}= 3.846D$

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Example Question #1 : Optics

A flower is placed 20cm away from a converging lens. An upside down image of the flower is measured as 6cm tall and at a distance of 30cm from the lens. Find the focal length of the lens and the real height of the flower.

Possible Answers:

f=20cm

$h_{o}=12cm$

f=0.08cm

$h_{o}=6.7cm$

f=12cm

$h_{o}=4cm$

f=30.5cm

$h_{o}=4.5cm$

Correct answer:

f=12cm

$h_{o}=4cm$

Explanation:

Let the object distance be $d_{0}$ and the image distance be $d_{i}$ . The object distance is the distance the flower is from the lens. The image distance is the distance the image is located from the lens. The thin lens equation is

$\frac{1}{f}=\frac{1}{d_{o}}+\frac{1}{d_{i}}$

$f=\left ( \frac{1}{d_{o}}+\frac{1}{d_{i}}\right )^{-1}=12cm$

Alternatively you can use

$f=\frac{d_{o}d_{i}}{d_{o}+d_{i}}$

To find the height, the equation for the magnification of an object is given by

$m=-\frac{d_{i}}{d_{o}}$

it is also given as

$m=-\frac{h_{o}}{h_{i}}$

The negative sign indicates an image that is upside down. Setting these equal gives

$\frac{d_{i}}{d_{o}}=\frac{h_{i}}{h_{o}} \rightarrow h_{o}=\frac{d_{o}}{d_{i}}h_{i}=4cm$

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Example Question #1 : Optics

You are passing a ray of light through a thin tank of alcohol to determine properties. Assume that the light ray is not effected by the tank's exterior. You find that if a piece of white paper is put underneath the alcohol tank, there is an image on the metal.

If the top of the image is 10cm from the center of the tank, and the top of the object is located 3mm from the center of the tank, what is the focal length of this lens?

Possible Answers:

f= 2.9

f=340

f=5.5

f= .0029

Correct answer:

f= .0029

Explanation:

Using object distance, image distance, and focal length formula:

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

Where is object distance, is image distance, and is focal length.

Plugging everything in:

$\frac{1}{.003}+\frac{1}{.1}=\frac{1}{f}$

Solving for ,

f= .0029

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AP Physics 2 : Thin Lens Equation

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Optics

Example Question #1 : Optics

Example Question #1 : Optics

All AP Physics 2 Resources

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