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33. Geometric Optics

Refraction At Spherical Surfaces

Physics

33. Geometric Optics

Refraction At Spherical Surfaces

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4:54 minutes

Problem 24

Textbook Question

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the of the arrow formed by paraxial rays incident on the convex surface. Is the erect or inverted?

Verified step by step guidance

Step 1: Identify the problem as one involving refraction at a spherical surface. Use the formula for refraction at a spherical surface:

n

₂s₂ − n₁s₁ = n₂ − n₁R, where

n

₁ and

n

₂ are the indices of refraction of the initial and final media,

s

₁ and

s

₂ are the object and image distances, and

R

is the radius of curvature of the surface.

Step 2: Assign the given values:

n

₁ = 1.00 (air),

n

₂ = 1.60 (glass),

s

₁ = −24.0 cm (negative because the object is on the same side as the incoming light), and

R = 4.00

cm (positive because the center of curvature is on the same side as the outgoing light). Substitute these values into the formula.

Step 3: Solve for

s

₂, the image distance. Rearrange the formula to isolate

s

₂:

s

₂ = n₂s₁Rn₁R + (n₂ − n₁)s₁. Substitute the known values into this equation.

Step 4: Determine the magnification of the image using the magnification formula:

m = − s

₂s₁. Use the value of

s

₂ obtained in the previous step and the given

s

₁ to calculate the magnification.

Step 5: Calculate the height of the image using the magnification:

h

_image = mh_object. Substitute the given object height (

h

_object = 1.50 mm) and the magnification obtained in the previous step. Finally, determine whether the image is erect or inverted based on the sign of the magnification (positive for erect, negative for inverted).

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Key Concepts

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

Refraction and the Index of Refraction

Refraction is the bending of light as it passes from one medium to another, which occurs due to a change in its speed. The index of refraction (n) quantifies this effect, defined as the ratio of the speed of light in a vacuum to its speed in the medium. For example, an index of refraction of 1.60 indicates that light travels 1.60 times slower in glass than in a vacuum, affecting how light rays converge or diverge when they hit a surface.

Convex Lens and Image Formation

A convex lens is thicker at the center than at the edges and converges incoming parallel light rays to a focal point. The position and characteristics of the image formed by a convex lens can be determined using the lens formula (1/f = 1/do + 1/di) and magnification equations. The nature of the image (real or virtual, erect or inverted) depends on the object's position relative to the focal length.

Paraxial Rays and Ray Diagrams

Paraxial rays are light rays that make small angles with the optical axis of a lens or mirror, allowing for simplified calculations in optics. Ray diagrams utilize these rays to trace the path of light through optical systems, helping to determine the position and size of images. In this context, analyzing paraxial rays is essential for predicting how the arrow's image will be formed by the convex surface of the rod.

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Multiple Choice

A student observes a goldfish in a flat sided aquarium. The goldfish observes the student. If the student observes that the image of the fish's nose is

6.0 cm

from the side of the tank, how far behind the glass is the nose of the fish?

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Multiple Choice

One side of a biconvex glass lens (n=1.5) has a radius of curvature of

25 cm

, while the other side has a radius of curvature of

40 cm

. What is its focal length?

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Multiple Choice

An object is embedded in glass as shown in the following figure. If the glass has a concave face, and is embedded in water, where will the image be located? Will the image be real or virtual?

A Spherical Fish Bowl. A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 cm in diameter. Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored.

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Textbook Question

A goldfish lives in a 50-cm-diameter spherical fish bowl. The fish sees a cat watching it. If the cat's face is 20 cm from the edge of the bowl, how far from the edge does the fish see it as being? (You can ignore the thin glass wall of the bowl.)

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Textbook Question

Paraxial light rays approach a transparent sphere parallel to an optical axis passing through the center of the sphere. The rays come to a focus on the far surface of the sphere. What is the sphere's index of refraction?

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Textbook Question

(a) A plane mirror can be considered a limiting case of a spherical mirror. Specify what this limit is. (b) Determine an equation that relates the image and object distances in this limit of a plane mirror.

303

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Textbook Question

(d) Are your results in parts (b) and (c) consistent with the discussion of Section 32–2 on plane mirrors?

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