<< 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!
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