<< Chapter < Page Chapter >> Page >

Refraction at a convex surface

Consider a point source of light at point P in front of a convex surface made of glass (see [link] ). Let R be the radius of curvature, n 1 be the refractive index of the medium in which object point P is located, and n 2 be the refractive index of the medium with the spherical surface. We want to know what happens as a result of refraction at this interface.

Figure shows a section of a sphere. The refractive index of air is n subscript 1 and that of the sphere is n subscript 2. Centre of the sphere is C and radius is R. A ray originating from point P on the optical axis outside the sphere strikes the convex surface of the sphere and is refracted within it. It intersects the axis at point P prime within the sphere, on the other side of the center. A dotted line labeled normal to interface connects the center of the sphere to the point of incidence. It makes an angle phi with the optical axis. The incident and refracted rays make angles alpha and beta respectively with the optical axis and angles theta 1 and theta 2 respectively with the normal to interface.
Refraction at a convex surface ( n 2 > n 1 ) .

Because of the symmetry involved, it is sufficient to examine rays in only one plane. The figure shows a ray of light that starts at the object point P , refracts at the interface, and goes through the image point P . We derive a formula relating the object distance d o , the image distance d i , and the radius of curvature R .

Applying Snell’s law to the ray emanating from point P gives n 1 sin θ 1 = n 2 sin θ 2 . We work in the small-angle approximation, so sin θ θ and Snell’s law then takes the form

n 1 θ 1 n 2 θ 2 .

From the geometry of the figure, we see that

θ 1 = α + ϕ , θ 2 = ϕ β .

Inserting these expressions into Snell’s law gives

n 1 ( α + ϕ ) n 2 ( ϕ β ) .

Using the diagram, we calculate the tangent of the angles α , β , and ϕ :

tan α h d o , tan β h d i , tan ϕ h R .

Again using the small-angle approximation, we find that tan θ θ , so the above relationships become

α h d o , β h d i , ϕ h R .

Putting these angles into Snell’s law gives

n 1 ( h d o + h R ) = n 2 ( h R h d i ) .

We can write this more conveniently as

n 1 d o + n 2 d i = n 2 n 1 R .

If the object is placed at a special point called the first focus , or the object focus F 1 , then the image is formed at infinity, as shown in part (a) of [link] .

Figure a shows a section of a sphere and a point F1 outside it, on the optical axis. Rays originating from F1 strike the convex surface and are refracted within the sphere as parallel rays. The distance of F1 from the surface is f subscript 1. Figure b shows rays parallel to the optical axis striking the convex surface and being refracted. They converge at point F2 within the sphere. F2 lies on the optical axis between the surface and the center of the sphere. The distance of F2 from the surface is f subscript 2. In both figures the refractive index of air is n1 and that of the sphere is n2 greater than n1.
(a) First focus (called the “object focus”) for refraction at a convex surface. (b) Second focus (called “image focus”) for refraction at a convex surface.

We can find the location f 1 of the first focus F 1 by setting d i = in the preceding equation.

n 1 f 1 + n 2 = n 2 n 1 R
f 1 = n 1 R n 2 n 1

Similarly, we can define a second focus or image focus F 2 where the image is formed for an object that is far away [part (b)]. The location of the second focus F 2 is obtained from [link] by setting d o = :

n 1 + n 2 f 2 = n 2 n 1 R
f 2 = n 2 R n 2 n 1 .

Note that the object focus is at a different distance from the vertex than the image focus because n 1 n 2 .

Sign convention for single refracting surfaces

Although we derived this equation for refraction at a convex surface, the same expression holds for a concave surface, provided we use the following sign convention:

  1. R > 0 if surface is convex toward object; otherwise, R < 0 .
  2. d i > 0 if image is real and on opposite side from the object; otherwise, d i < 0 .

Summary

This section explains how a single refracting interface forms images.

  • When an object is observed through a plane interface between two media, then it appears at an apparent distance h i that differs from the actual distance h o : h i = ( n 2 / n 1 ) h o .
  • An image is formed by the refraction of light at a spherical interface between two media of indices of refraction n 1 and n 2 .
  • Image distance depends on the radius of curvature of the interface, location of the object, and the indices of refraction of the media.

Conceptual questions

Derive the formula for the apparent depth of a fish in a fish tank using Snell’s law.

Got questions? Get instant answers now!

Use a ruler and a protractor to find the image by refraction in the following cases. Assume an air-glass interface. Use a refractive index of 1 for air and of 1.5 for glass. ( Hint : Use Snell’s law at the interface.)

(a) A point object located on the axis of a concave interface located at a point within the focal length from the vertex.

(b) A point object located on the axis of a concave interface located at a point farther than the focal length from the vertex.

(c) A point object located on the axis of a convex interface located at a point within the focal length from the vertex.

(d) A point object located on the axis of a convex interface located at a point farther than the focal length from the vertex.

(e) Repeat (a)–(d) for a point object off the axis.

answers may vary

Got questions? Get instant answers now!

Problems

An object is located in air 30 cm from the vertex of a concave surface made of glass with a radius of curvature 10 cm. Where does the image by refraction form and what is its magnification? Use n air = 1 and n glass = 1.5 .

Got questions? Get instant answers now!

An object is located in air 30 cm from the vertex of a convex surface made of glass with a radius of curvature 80 cm. Where does the image by refraction form and what is its magnification?

d i = −55 cm ; m = + 1.8

Got questions? Get instant answers now!

An object is located in water 15 cm from the vertex of a concave surface made of glass with a radius of curvature 10 cm. Where does the image by refraction form and what is its magnification? Use n water = 4 / 3 and n glass = 1.5 .

Got questions? Get instant answers now!

An object is located in water 30 cm from the vertex of a convex surface made of Plexiglas with a radius of curvature of 80 cm. Where does the image form by refraction and what is its magnification? n water = 4 / 3 and n Plexiglas = 1.65 .

d i = −41 cm, m = 1.4

Got questions? Get instant answers now!

An object is located in air 5 cm from the vertex of a concave surface made of glass with a radius of curvature 20 cm. Where does the image form by refraction and what is its magnification? Use n air = 1 and n glass = 1.5 .

Got questions? Get instant answers now!

Derive the spherical interface equation for refraction at a concave surface. ( Hint : Follow the derivation in the text for the convex surface.)

proof

Got questions? Get instant answers now!

Questions & Answers

what does preconceived mean
sammie Reply
physiological Psychology
Nwosu Reply
How can I develope my cognitive domain
Amanyire Reply
why is communication effective
Dakolo Reply
Communication is effective because it allows individuals to share ideas, thoughts, and information with others.
effective communication can lead to improved outcomes in various settings, including personal relationships, business environments, and educational settings. By communicating effectively, individuals can negotiate effectively, solve problems collaboratively, and work towards common goals.
it starts up serve and return practice/assessments.it helps find voice talking therapy also assessments through relaxed conversation.
miss
Every time someone flushes a toilet in the apartment building, the person begins to jumb back automatically after hearing the flush, before the water temperature changes. Identify the types of learning, if it is classical conditioning identify the NS, UCS, CS and CR. If it is operant conditioning, identify the type of consequence positive reinforcement, negative reinforcement or punishment
Wekolamo Reply
please i need answer
Wekolamo
because it helps many people around the world to understand how to interact with other people and understand them well, for example at work (job).
Manix Reply
Agreed 👍 There are many parts of our brains and behaviors, we really need to get to know. Blessings for everyone and happy Sunday!
ARC
A child is a member of community not society elucidate ?
JESSY Reply
Isn't practices worldwide, be it psychology, be it science. isn't much just a false belief of control over something the mind cannot truly comprehend?
Simon Reply
compare and contrast skinner's perspective on personality development on freud
namakula Reply
Skinner skipped the whole unconscious phenomenon and rather emphasized on classical conditioning
war
explain how nature and nurture affect the development and later the productivity of an individual.
Amesalu Reply
nature is an hereditary factor while nurture is an environmental factor which constitute an individual personality. so if an individual's parent has a deviant behavior and was also brought up in an deviant environment, observation of the behavior and the inborn trait we make the individual deviant.
Samuel
I am taking this course because I am hoping that I could somehow learn more about my chosen field of interest and due to the fact that being a PsyD really ignites my passion as an individual the more I hope to learn about developing and literally explore the complexity of my critical thinking skills
Zyryn Reply
good👍
Jonathan
and having a good philosophy of the world is like a sandwich and a peanut butter 👍
Jonathan
generally amnesi how long yrs memory loss
Kelu Reply
interpersonal relationships
Abdulfatai Reply
What would be the best educational aid(s) for gifted kids/savants?
Heidi Reply
treat them normal, if they want help then give them. that will make everyone happy
Saurabh
What are the treatment for autism?
Magret Reply
hello. autism is a umbrella term. autistic kids have different disorder overlapping. for example. a kid may show symptoms of ADHD and also learning disabilities. before treatment please make sure the kid doesn't have physical disabilities like hearing..vision..speech problem. sometimes these
Jharna
continue.. sometimes due to these physical problems..the diagnosis may be misdiagnosed. treatment for autism. well it depends on the severity. since autistic kids have problems in communicating and adopting to the environment.. it's best to expose the child in situations where the child
Jharna
child interact with other kids under doc supervision. play therapy. speech therapy. Engaging in different activities that activate most parts of the brain.. like drawing..painting. matching color board game. string and beads game. the more you interact with the child the more effective
Jharna
results you'll get.. please consult a therapist to know what suits best on your child. and last as a parent. I know sometimes it's overwhelming to guide a special kid. but trust the process and be strong and patient as a parent.
Jharna
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 3

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 3' conversation and receive update notifications?

Ask