Physics - Optics: Refracted Surface (5 of 5)

Michel van Biezen
6 May 201303:24
EducationalLearning
32 Likes 10 Comments

TLDRThis educational script demonstrates the process of finding the image of an object through a refracting surface, such as a glass lens. It explains the use of Snell's law with given indices of refraction, object distance, and radius of curvature to calculate the image distance. The example concludes with the determination of a virtual image's position and magnification, revealing an upright, half-sized image 15 centimeters from the surface, emphasizing the principles of light refraction and image formation.

Takeaways
  • 🔍 The script explains the process of finding the image of an object when light rays cross a refracting surface.
  • 🌐 The object is in air and the observer is on the opposite side of the boundary.
  • 🧭 The index of refraction on the object side is 1, and on the other side, it's 1.5, likely representing glass.
  • 📏 The radius of curvature of the refracting surface is given as -10 centimeters, indicating it bends towards the observer.
  • 📘 The lens formula used is \( \frac{n_1}{s} + \frac{n_2}{s'} = \frac{n_2 - n_1}{r} \), where \( s \) is the object distance, \( s' \) is the image distance, and \( r \) is the radius of curvature.
  • ✍️ The values are substituted into the lens formula to find \( s' \), the distance to the image.
  • 🔢 Simplification of the equation leads to finding \( s' \) as -15 centimeters, indicating a virtual image.
  • 🔄 The negative sign of \( s' \) means the image is on the same side as the object, 15 centimeters from the surface.
  • 🔎 To determine the size of the image, the magnification equation is used, which is \( m = -\frac{n_1 s'}{n_2 s} \).
  • 📏 The calculated magnification is one half, meaning the image is upright and half the size of the object.
  • 📚 The script concludes by stating that the image is virtual, upright, half the size of the object, and located 15 centimeters from the refracting surface.
Q & A

  • What is the index of refraction on the object side?

    -The index of refraction on the object side (n1) is 1.

  • What is the index of refraction on the observer's side?

    -The index of refraction on the observer's side (n2) is 1.5, which is probably glass.

  • What is the radius of curvature of the refracting surface?

    -The radius of curvature of the refracting surface is 10 centimeters, and it is bent towards the observer, making it a negative value.

  • What is the object distance (s) in the problem?

    -The object distance (s) is 20 centimeters.

  • How is the radius of curvature represented in the equation?

    -The radius of curvature (r) is represented as a negative number because it is bent towards the observer, making it -10.

  • What equation is used to find the image distance (s')?

    -The equation used is n1/s + n2/s' = (n2 - n1)/r.

  • What is the calculated image distance (s')?

    -The calculated image distance (s') is -15 centimeters.

  • What does the negative sign in the image distance indicate?

    -The negative sign indicates that the image is virtual and on the same side as the object.

  • How is the magnification of the image determined?

    -The magnification of the image is determined using the equation magnification = -n1 * s' / (n2 * s).

  • What is the magnification and orientation of the image?

    -The magnification is 0.5, meaning the image is half the size of the object, and since it's a positive magnification, the image is upright.

Outlines
00:00
🔍 Finding the Image in Refraction

The script explains the process of determining the image position and magnification when an object is viewed through a refracting surface, such as glass, with a negative radius of curvature. The object is in air, and the observer is on the other side of the boundary. The index of refraction on the object side is 1, and on the other side, it is 1.5. Using Snell's law, the script demonstrates the calculation of the image distance (s') and magnification. The radius of curvature is given as -10 cm, and the object distance is 20 cm. The calculation results in a virtual image located 15 cm from the surface, half the size of the object, and upright.

Mindmap
Keywords
💡Refraction
Refraction is the change in direction of a wave passing from one medium to another, due to a change in speed. In the video, refraction is the central concept as it explains how light rays change direction when passing from air to glass, which is crucial for understanding the formation of the image.
💡Index of Refraction
The index of refraction is a measure of how much light slows down and bends when it enters a medium. In the script, the index of refraction is used to calculate the bending of light rays at the boundary between air and glass, with the values 1 for air and 1.5 for glass being key to the calculations.
💡Object Distance
Object distance refers to the distance between the object and the refracting surface. In the video, the object is placed 20 centimeters away from the glass surface, which is an essential measurement for determining the position of the image.
💡Image Distance
Image distance is the distance from the refracting surface to the image formed by the refraction of light. The script focuses on calculating this distance, which is found to be 15 centimeters on the same side as the object, indicating a virtual image.
💡Radius of Curvature
The radius of curvature is the distance from the center of curvature of a lens or mirror to its surface. In the script, the lens has a radius of curvature of -10 centimeters, indicating it curves towards the observer, which affects the refraction of light.
💡Virtual Image
A virtual image is formed when the light rays appear to diverge from a point behind the refracting surface. The script explains that the image formed is virtual, as it is located 15 centimeters in front of the surface on the same side as the object.
💡Lensing Equation
The lensing equation, or the thin lens equation, is used to calculate the image distance based on the object distance and the lens properties. The script uses a variation of this equation to find the image distance after light passes through the refracting surface.
💡Magnification
Magnification is the ratio of the size of the image to the size of the object. The script calculates the magnification of the image to be half the size of the object, indicating that the image is smaller and upright.
💡Upright Image
An upright image is one where the orientation of the image is the same as that of the object. The script mentions that the magnification is positive, which means the image is upright and in the same direction as the object.
💡Negative Radius of Curvature
A negative radius of curvature indicates that the surface curves towards the observer. In the script, this is used to determine the direction and properties of the refraction, leading to the formation of a virtual image.
💡Simplification
Simplification in the context of the script refers to the mathematical process of making equations easier to solve. The script simplifies the lensing equation to find the image distance more conveniently, which is a crucial step in the problem-solving process.
Highlights

Demonstration of finding the image of an object when light rays pass through a refracting surface.

Object is in air and observer is on the opposite side of the boundary.

Index of refraction on the object side is 1, while the other side is 1.5, likely glass.

Radius of curvature is 10 cm, but negative due to bending towards the observer.

Use of lens formula n1/s + n2/s' = n2 - n1/r to find the image.

Plugging in the numbers to solve for the image distance s'.

Simplification of the lens formula to find s' = -15 cm, indicating a virtual image.

Explanation of the negative sign meaning the image is on the same side as the object.

Virtual image located approximately 15 cm from the refracting surface.

Use of magnification equation to determine the size of the image.

Calculation of magnification as half the size of the object.

Positive magnification indicating the image is upright and in the same direction as the object.

Placement of the image upright and half the size, 15 cm away from the surface.

Clarification that s' = -15 cm means the image is in front of the surface, not behind.

Virtual image characteristics explained in the context of the problem.

Step-by-step problem-solving approach to finding and characterizing the image.

Transcripts

Browse More Related Video

Physics - Optics: Refracted Surface (4 of 5) Glass Ball

Physics 54 Optics: Mirrors (4 of 6) Convex Mirror

Physics 51/54 - Optics: Mirrors (6 of 6) Flat Mirror

Physics - Optics: Lenses (2 of 2) Diverging Lens

Physics - Optics: Lenses (1 of 2) Diverging Lens

Physics - Optics: Lenses (2 of 5) Lens Combinations - Two Converging Lenses

Rate This

5.0 / 5 (0 votes)

Thanks for rating:

Related Tags
RefractionVirtual ImageOpticsPhysicsLight RaysIndex of RefractionCurvatureMagnificationOptical PhenomenaEducational