The biggest problem in dealing with lenses and mirrors is to keep the sign convention straight. It varies from book to book and from teacher to teacher. The convention used here will be the following: 


d₀ and dᵢare + if they are in front of the mirror. d₀and dᵢ are - if they are in back of the mirror. do means distance of the object from the mirror and di means distance of the image from the mirror.

f (focal point) is + if the center of curvature is in front of the mirror and - if the center of curvature is behind the mirror. The same sign convention is used for R (radius of curvature).

Magnification (M) of a lens is the number multiplied by the object size to get the image size.  M is + if the image is erect. M is - if the image is inverted. 


d₀ is + if the object is in front of the lens and d₀  is - if the object is behind the lens.

dᵢ is + if the image is behind the lens. dᵢ is - if the image is in front of the lens.

f is + for a converging lens and - for a diverging lens.


Plane mirrors form images that are erect, the same size as the object, and as far behind the reflecting surface as the object is in front of it. This type of image is called virtual. A virtual image will not appear on a screen located at the position on the image because the light does not converge there. The eye (or camera) interprets the diverging rays as coming from an image behind the mirror! 

The principal focus of a spherical mirror is the point F where rays parallel to and very close to the central or optical axis of the mirror are focused. This focus is real for a concave mirror and virtual for a convex mirror. It is located on the optical axis and midway between the center of curvature C and the mirror. Concave mirrors form inverted real images of objects placed beyond the principal focus. If the object is between the principal focus and the mirror, the image is virtual, erect, and enlarged. Convex mirrors produce only erect virtual images of objects placed in front of them. The images are diminished (smaller than the object) in size.

The mirror equation for both concave and convex spherical mirrors is:


where d₀ = distance of the object from the mirror; dᵢ = distance of the image from the mirror;
f = focal point.

   The size of the image form by a spherical mirror is given by:

Linear magnification = length of image/length of object = dᵢ/d₀


Converging (negative) lenses, or convex lenses, are thicker at the center than at the rim and will converge a beam of parallel light to a real focus.

Diverging (positive) lenses, or concave lenses, are thinner at the center than at the rim and will diverge a beam of parallel light from a virtual focus.

The principal focus (focal point) of a thin lens with spherical surfaces is the point F where rays parallel to and near the central or optical axis are brought to a focus; this focus is real for a converging lens and virtual for a diverging lens.

The object and image relation for converging and diverging lenses is:

where d₀ = object distance from the lens; dᵢ = image distance from the lens; f = focal length of the lens. The lens is assumed to be thin, and the light rays paraxial (close to the principal focus). 

Linear image = size of image/size of object = image distance from lens/object distance from lens = dᵢ/d₀

Converging lenses form inverted real images of objects located outside the principal focus.

When the object is between the principal focus and the lens, the image is virtual (on the same side of the lens as the object), erect, and enlarged. 


Diverging lenses produce only virtual, erect, and smaller images of real objects.

The Lensmaker's Equation which holds for all types of thin lenses is

1/f = (n-1) (1/r₁ - 1/r₂)

where n = refractive index of the lens material; r1 and r2 are the radii of curvature of the two lens surfaces. This equation holds for all types of thin lenses. A radius of curvature, r, is positive when its center of curvature lies to the right of the surface, and negative when its center of curvature lies to the left of the surface.

Lens power in diopters (m-¹) is equal to 1/f, where f is the focal length expressed in meters.

When two lenses having focal lengths f₁ and f₂ are in close contact, the focal length f of the combination is

1/f = 1/f₁ + 1/f₂

For lenses in close contact, the power of the combination is equal to the sum of their individual powers.


Diffraction and Interference

As mentioned in earlier lessons, Huygens considered light to be a wave. He described a wave crest advancing by imagining each point along the wave crest to be a source point for small, circular, expanding wavelets, which expand with the speed of the wave. The surface tangent to these wavelets determines the contour of the advancing wave. Huygens principle can be used to derive the law of reflection and the law of refraction. The observed laws of geometric optics follow from the assumption that light is a wave.


Because light is a wave, the superposition principle is valid to determine the constructive and destructive interference for light waves. For constructive interference, two waves must have the two contributing waves and the two troughs arriving at the same time. For destructive interference, a crest from one wave and a trough from the other must arrive at a given point at the same time.

 Thomas Young first demonstrated interference from light waves with a double slit experiment. Each of the slits acts as a source for circular expanding waves. The points of intersection of two crests, one from each slit, are points of constructive interference. The point of intersection of a crest from one slit and a trough from the other slit is a point of destructive interference. Therefore, the interference pattern called fringes, consisting of alternating light and dark bars, will be seen on the screenYoung's double slit experiment shows that light spreads out in wavefronts that can interfere with each other. Diffraction is the effect of a wave spreading as it passes through an opening or goes around an object. Diffraction of sound is quite obvious however, diffraction of light is difficult to observe. Diffraction is proportional to wavelength, so it is not easy to observe the bending of light when it passes through a small aperture or goes around a sharp edge.

   When looking through double slits, it is impossible to see only the double-slit pattern because the double slit is really two single slits; therefore, the actual observed pattern is that of superimposed double and single slit patterns.

For double slit diffraction, the condition for constructive interference (or bright fringes) is

 dsin2 = m8

where d is the distance between slits; m is an integer (0...1...2...3...etc.). The angles 2 are where the light has maximum intensity

The condition for destructive interference in double slit diffraction is

dsin2 = (m + 1/2)8

For single slit diffraction, the condition for destructive interference is 

sin2 = m8/"

where " is the width of the single slit.

When light enters a medium with a refractive index n, the speed of light is divided by n and the wavelength also gets divided by n. The frequency remains the same. 

Review Questions:

1. An object is 4.0 cm in front of a plane mirror. Where does the observer looking into the mirror see the image? Is it real or virtual?

Image is 4.0 cm behind the mirror and it is virtual.   

2. For a spherical concave mirror of 12 cm radius of curvature, describe the image of a 2.0 cm height object placed 20 cm on the center line of the mirror.

The image is 8.6 cm from the mirror, inverted and real.   

3. For a spherical convex mirror of 14 cm radius of curvature, describe the image of a 2.5 cm object placed 30 cm out on the principal axis of the mirror.

  Magnification is 0.19; image height is 0.48 cm; image is erect and virtual.   

4. A converging lens of focal length 8.0 cm forms an image of an object placed 20 cm in front of the lens. Describe the image.

 The image is 13 cm on the side of the lens opposite the object with magnification 0.67. It is inverted and real.   

5. A diverging lens has a 14 cm focal length. Describe the image of a 4.0 cm object placed 40 cm from the lens. Describe the image.

The image is 1.0 cm high, erect, virtual, and appears to come from a point 10 cm from the lens on the same side as the object.   

6. An object that is 0.87 cm tall and 4.9 cm from a convex lens produces a real image that is 2.5 cm tall from the other side of the lens. (a) What is the size of the image? Is the image erect or inverted? Is the image enlarged or reduced? (b) What is the magnification of the lens? What is the focal length of the lens?

a. d = 3071 nm
b. 11.2 degrees
c. 17 degrees 

7. Light with a wavelength of 600 nm falls on a double slit diffraction setup and the second bright fringe is seen at an angle of 23°. (a) Find the distance between the slits. (b) Find the angle for the first bright fringe and (c) find the angle for the m = 1 dark fringe.

a. d = 3071 nm
b. 11.2 degrees
c. 17 degrees

8. Waves are broadcast from a 1800 kHz radio station which is 1 km from a home. The radio transmitter is at the same height as the home. The home is in the path of an airport, and when planes fly over head, the signal is sometimes lost when a plane passes directly overhead. Find the first minimum height of the plane where the condition for destructive interference is satisfied at the home.

80.1 m 

9. Monochromatic light is incident on a single 4.0 x 10-4 m wide slit producing a diffraction pattern on a screen 1.5 m away. The distance from the central maxima to the second dark fringe is 3.4 x 10-3 m. What is the wavelength of the light?

453 nm 

10. A grating with 2000 lines per cm is illuminated with a hydrogen gas discharge tube. Two of the hydrogen lines are at 410 and 434 nm. What is the first order spacing of these lines on a screen 1.0 m from the grating?

The lines arte separated by 5 mm on the screen.   


11. A convex mirror has a focal length of -12 cm. A lightbulb with a diameter of 6.0 cm is placed 60.0 cm in front of the mirror. Locate the image of the lightbulb. What is the diameter of the image?

  1.0 cm 

12. You used a convex lens to produce an image with di = 25 cm and hi = 4.0 cm. You place a concave lens with a focal length of -15 cm between the convex lens and the original image. To your surprise you see a real, enlarged image on the wall. You are told that the image from the convex lens is now the object for the concave lens, and because it is on the opposite side of the concave lens, it is a virtual lens. Find the location and size of the new image and predict whether the concave lens changed the orientation of the original image.

image distance = +3.0 cm; M = +3; image height = 10 cm 

13. Lynn uses one of her mom's old 33-1/3 rpm records as a diffraction grating. She shines her laser, with wavelength of 632.8 nm on the record. on a screen 4.0 m from the record, a series of red dots 21 mm apart are visible. (a) How many ridges are there in a centimeter along the radius of the record? (b) She checks her results by noting that the ridges came from a song that lasted 4.01 minutes and took 16 mm on the record. How many ridges should be in a centimeter? 

a. 83 ridges/cm
b. 84 ridges/cm