AP Physics 2 : Thin Lens Equation

Study concepts, example questions & explanations for AP Physics 2

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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  from his eyes to the book. His near-point distance is . If his glasses are  from his eyes, what is the refractive power of his glasses lenses?

Possible Answers:

Correct answer:

Explanation:

Write the thin-lens equation.

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

The image distance is:

Substitute the givens to the equation.

Refractive power is the inverse of focal length in .

Example Question #1 : Thin Lens Equation

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

Possible Answers:

Correct answer:

Explanation:

Let the object distance be  and the image distance be . 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

Alternatively you can use

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

it is also given as

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

Example Question #1 : Thin Lens Equation

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  from the center of the tank, and the top of the object is located  from the center of the tank, what is the focal length of this lens? 

Possible Answers:

Correct answer:

Explanation:

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

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

Plugging everything in:

Solving for ,

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