A fortune teller's 'crystal ball' (actually just glass) is 10 cm in diameter. Her secret ring is placed 6.0 cm from the edge of the ball. The crystal ball is removed and a thin lens is placed where the center of the ball had been. If the image is still in the same position, what is the focal length of the lens?

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Determine the radius of the crystal ball. Since the diameter is given as 10 cm, the radius is half of that: \( r = \frac{10}{2} = 5 \, \text{cm} \).

Understand that the crystal ball acts as a spherical lens. The distance of the object (the ring) from the center of the ball is \( d_o = 6.0 + 5 = 11.0 \, \text{cm} \), where 6.0 cm is the distance from the edge of the ball to the ring, and 5 cm is the radius of the ball.

The image formed by the crystal ball is at the same position as the object distance relative to the center of the ball. This means the image distance \( d_i \) is also \( 11.0 \, \text{cm} \).

When the crystal ball is replaced by a thin lens, the lens equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) can be used to find the focal length \( f \). Substitute \( d_o = 11.0 \, \text{cm} \) and \( d_i = 11.0 \, \text{cm} \) into the equation.

Simplify the lens equation: \( \frac{1}{f} = \frac{1}{11.0} + \frac{1}{11.0} \). Solve for \( f \) to find the focal length of the lens.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lens Formula

The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens. It is expressed as 1/f = 1/v - 1/u. This formula is essential for determining the focal length of a lens when the object and image distances are known.

Sign Convention in Optics

In optics, the sign convention helps in determining the signs of distances in the lens formula. Typically, distances measured in the direction of the incoming light are negative, while those in the direction of outgoing light are positive. Understanding this convention is crucial for correctly applying the lens formula.

Thin Lens Approximation

The thin lens approximation assumes that the thickness of the lens is negligible compared to the object and image distances. This simplification allows for easier calculations using the lens formula, making it applicable in many practical scenarios, such as the one described in the question.

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Thin Lens Equation

Related Practice

Textbook Question

CALC FIGURE CP34.81 shows a light ray that travels from point A to point B. The ray crosses the boundary at position x, making angles θ₁ and θ₂ in the two media. Suppose that you did not know Snell's law. You've proven that Snell's law is equivalent to the statement that 'light traveling between two points follows the path that requires the shortest time.' This interesting way of thinking about refraction is called Fermat's principle. Write an expression for the time t it takes the light ray to travel from A to B. Your expression should be in terms of the distances a, b, and w; the variable x; and the indices of refraction n₁ and n₂.

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Textbook Question

A fortune teller's 'crystal ball' (actually just glass) is 10 cm in diameter. Her secret ring is placed 6.0 cm from the edge of the ball. An image of the ring appears on the opposite side of the crystal ball. How far is the image from the center of the ball?

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Textbook Question

Consider a lens having index of refraction n₂ and surfaces with radii R₁ and R₂. The lens is immersed in a fluid that has index of refraction n₁. A symmetric converging glass lens (i.e., two equally curved surfaces) has two surfaces with radii of 40 cm. Find the focal length of this lens in air and the focal length of this lens in water.

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