Chapter 1. Geometrical Optics
PG1. Single convex Surface.(see p.22)
A rod of material with refractive index n2 =1.5 has on the side facing the incident
light a convex spherical surface with radius of curvature of r = 50 cm.
a. What is the object distance to have the image at +7 cm ?
b. What is the object distance to have the image at -7 cm ?
c. Assuming r = 25 cm, make a graph of xi as function of xo for n1 = 1, n2 = 1.33 and
do the graphical construction of the image for
i) real objects before and after the object focal point, and
ii) virtual objects before and after the image focal point.
PG2. Single concave Surface.(see p.22)
A rod of material with refractive index n2 = 1.5 has on one side a concave spherical
surface with radius of curvature of r = -50 cm.
a. What is the object distance to have the image at +5cm ?
b. What is the object distance to have the image at -5 cm ?
c. Assuming for r = -25cm, make a graph of xi as function of xo for n1 = 1, n2 = 1.33
and do the graphical construction of the image for
i) real objects before and after the image focal point, and
ii) virtual objects before and after the object focal point.
PG3. Rod Sticks in Water, calculation of Image Distance.(see p.22)
A plastic rod of length 70cm is stuck vertically in water. An object is positioned on
the cross section at the top of the rod, which sticks out of the water and faces the sun.
On the other side in the water, the rod has a concave spherical surface, with respect to
the incident light from the sun, with r = - 4cm. The refractive index of the rod is n1
= 1.5, and of the water n2 = 1.33. Calculate the image distance of the object.
Air
n=1.5
water, n=1.33
70cm
r =-4cm
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PG4. Plastic Film on Water as Spherical Surface.(see p 22)
A plastic film is mounted on a ring and placed on the surface of water. The film forms
a spherical surface filled with water. The thickness of the film is neglected and
therefore we have a convex surface of water of n2 = 1.33. Sun light is incident on the
surface and the image is observed at 100 cm deep in the water. Calculate the radius of
curvature of the “spherical water surface”.
Air
water, n =1.33
100 cm
Image
PG5. Air Lens in Plastic.(see p.35)
A plastic rod is flat on one side and has a spherical surface on the other side. The
spherical surface has radius of curvature r = -10cm, it is concave with respect to the
incident light, which comes from the flat side. An identical second rod is taken and
the two curved ends are put together, forming an air lens by the ends of the two rods.
The cross section of this lens has its thinnest point in the middle. Assume that the
radius of curvatures of the second spherical surfaces is r’ = 10cm and the refractive
index of the rod is n = 1.5. Sun light is incident on an object on the face of the first
rod at 20 cm from the “air lens”. Find the image distance.
r =-10cm
Air
r = 10cm
n = 1.5
xo = -20cm
PG6. Positive thin Lens on Water.(see p.35)
A lens in air with radii of curvature r = 10cm and r’ = -10cm and refractive index n =
1.5 has a focal length of f = 10cm. The lens is put on water, one surface in air, the
other in water. The refractive index of water is n = 1.33. Sun light is incident on the
lens.
a. Calculate the distance of the image point from the lens.
b. What is the distance if a Silicon lens with n = 3.4 is applied.
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PG7. Magnifier.(see p.47)
A magnifier lens of f1 = 12 is placed 8 cm from the eye.
Find the position of xo1 for
a. The near point configuration, calculate the magnification.
b. The “infinite” configuration, calculate the angular magnification.
PG8. Microscope (Near Point).(see p.48)
A microscope has a first lens (objective) with focal length 0.31 cm and a magnifier
(ocular) lens of 1.79 cm. We assume that the image of the first lens is at 16 +f1 cm
(from the objective lens) and the eye is in the “near Point” configuration.
a. Find object distance for objective lens.
b. Find distance from first image and magnifier lens.
c. Find the distance between the lenses.
d. Find the Magnification.
e. Show that there is little difference in the results, if one uses 16 or 16 +f1.
PG9. Microscope (negative infinity).(see p.48)
A microscope has a first lens (objective) with focal length 0.31 cm and a magnifier
(ocular) lens of 1.79 cm. We assume that the image of the first lens is at 16 +f1 cm
(from the objective lens) and the eye is relaxed, looking at -infinity.
a. Find objective distance for objective lens.
b. Find distance from first image and magnifier lens.
c. Find the distance between the lenses.
d. Find the Magnification.
e. How large is the difference when using 16 or 16 +f1.
PG10. Kepler Telescope.(see p.48)
Make a suggestion for construction of a Kepler telescope with magnification of 4 and
10. At what higher number does the construction become unrealistic? Why ?
PG11. Galilean Telescope.(see p.48)
A Galilean Telescope has for the first lens f1 = 30 cm and for the negative lens f2 = 9.9 cm. If xo1 is large and the distance a between the two lenses is 20 cm, calculate
xi2, the image distance with respect to the negative lens. Calculate the magnification
and show that for the object at - infinity, one has again M = -f1/f2. The distance
between the two lenses is then f1+f2.
PG12. Laser Beam Expander.(see p.48)
A laser beam of diameter of 2mm should be expanded to a beam of 20 mm.
a. A biconvex and a biconcave lens should be used. The beam passes first the
biconcave lens of focal length –5 mm. At what distance from the first lens should one
place a biconvex lens of focal length 50 mm ?
b. Two biconvex lenses should be used, one with f = 5 mm, the other with f = 50 mm.
What is the distance from the first to the second lens.
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PG13. An Exercise for matrix Multiplication.(see p.65)
Draw two cartesian coordinate systems x,y and x’,y’, the second rotated by the angle
 with respect to the first. Identify the matrix
A
=
 cos 

 -sin 
sin  

cos  
with the rotation of x,y into x’, y’.
This is a rotation in the mathematical positive sense.
a. The matrix for rotation in the mathematical negative sense is obtained by
substituting for  the negative value -. Write the matrix A-1 and compare.
b. Show that A A-1 is the unit matrix.
c. The transposed matrix AT is obtained from A by interchanging the 2,1 and 1,2
elements. Show that in our case AT is equal to A-1 and A AT is the unit matrix.
d. Write the matrix for rotation by the angle 2 and call it B. Show that B =A A
PG14. Non commutation of Matrices.(see p.66)
In general two matrices A and B may not be commuted, that is AB is not equal to BA.
We will see this on the following example for a different sequence of the same
matrices.
We consider two hemispherical thick lenses where light is coming from the left. The
light hits the first lens L1 at a spherical surface of radius of curvature r, then traverses
the thickness d, and emerges from a plane surface. The second lens L2 has the
reversed order, first the plane surface, then thickness d, and then the curved surface
with the same radius of curvature r. The refractive indices of the lenses are n2 and
outside we assume n1 = n3 = 1.
Make a sketch. See how the two lenses are “different”. The product matrix for lens 1
and lens 2 are different for the two cases. Compare the position of the principal
planes. Compare for the case that r = .
PG15. System with Focal Length f.(see p.66)
Calculate, “by hand”, using the matrix method, the position of the two principal
planes for a system of two thin lenses, both of focal length f, and at distance f. Make
a sketch of the position of the principal planes and indicate the object and image
distance for the case that they have the same absolute value.
PG16. Convex-concave Lens.(see p.66)
The first surface has radius of curvature r1 = 20 cm, the
second radius of curvature r2 = -10cm with thickness of d =
5 cm and n = 1.5.
a. Calculate the principal planes and focal length and find
the image of an object positioned at 5 cm to the left of the
first surface.
b. Find the same result by using twice the imaging equation
of a single surface.
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PG17. Plane-Concave lens.
Consider a Plane-Concave lens of refractive index n =1.5,
thickness d = 2cm and radius of curvature r = -30cm. The
index on the outside is 1 and an object is located 12 cm
to the left of the first surface
a. Use the matrix method to find the image distance from V2. Is
the image real or virtual.
b. Assume parallel light from the left and find the focal point
distance from V2.
PG18. Convex-Concave Lens.
Consider a convex-concave lens of refractive index n =1.5, thickness d = 5cm and
radii of curvature r1 = 20cm and r2 = -10cm. An object is located 18 cm to the left of
the first surface.
n1 =1
n2 =1.5
n1 =1
r1
18cm
V1
r2
d
V2
a. Use the matrix method to find the image distance from V2. Is the image real or
virtual?
b. Assume parallel light from the left and find the focal point distance from V2.
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PG19. Comparison of Plane-Concave and Convex-Plane Lens.(see p.66)
Consider a plane-Concave and a Convex-plane Lens.
n=1
n=1
r
n = 1.5
d
V1
V2
n=1
n=1
r
n = 1.5
d
V1
V2
Assume for the thickness d = 4cm and the absolute value of the radius of curvature
equal 10cm. An object is placed 100 cm to the left of the lenses measured from V1.
Calculate for both lenses the focal length and image position measured from V2.
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PG20.Convex-Concave Lens (equal xov, xiv).
Consider a convex-concave lens of refractive index n =1.7, thickness d = 2cm and
radii of curvature of r1 = 60cm and r2 = -20cm.
n1 =1
n2 =1.5
n1 =1
r1
xov
r2
V1
d
V2
xiv
Use the matrix method to find the parameters of the lens such that xov, the distance of
the object from V1, is equal to the image distance xiv from V2. Compare with a thin
lens, where xov = xiv =2f..
PG21. Glass Sphere(see p.58)
Compare two spheres, a glass sphere with refractive index n =1.5 and a Silicon sphere
with refractive index 3.4, both with a radius 10 cm. An object is located 120 cm to the
left of the sphere (measured from the first refracting surface).
r
n = 1.5
a. Use the matrix method to calculate the image distance.
b. Assume that the thin-lens equation can be used. Solve for the final image point
position, using the center of the sphere as reference.
c. Compare the results of a. and b. and give the error.
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PG22. Short Focal Length Lens.
Consider a convex-plan lens of refractive index n =1.5 and thickness d = 5cm. The
lens is used to concentrate parallel light on a detector, and the detector is placed at the
second surface of the lens.
n1 =1
n2 =1.5
n1 =1
r
parallel light
V1
d
V2
What is the radius of curvature of the first surface?.
PG23. Thick Concentration Lens.(see p.66)
A thick lens of radius of curvature r1 = 5 mm and r2 = -5 mm, refractive index n = 1.5,
and thickness 4 mm is used to concentrate incident parallel light on a detector. Using
the matrix method, find the position of the lens with respect to the detector plane.
PG24. Surface Corrections of hot Laser Rod.
A laser rod of length 20cm has at room temperatures two flat surfaces. When the rod
is in operation, it heats up and the flat surfaces show bending with radii of curvature
r1 = -r2. To determine these radii, parallel light enters the rod from the left and the
distance of the image point from the vertex V2 is measured to be 50cm.
n1 =1
n2 =1.5
r1
Parallel light
n1 =1
r2
V1
(image) F point
d=20
V2
q=50
Determine the radii of curvature of the hot laser rod.
After the radii r1 and r2 have been determined, the flat surfaces of the rod are
reshaped as spherical surfaces of radii negative r1 and positive r2. When in
operation, the surfaces of the rod bend again and are now nearly flat during
operation.
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PG25.Laser burning of Image
A thin lens of refractive index n =1.5 and radii of curvature r1 and r2 is used with
laser light to burn a figure into a plane in a medium of refractive index n = 1.6.
n1 =1
n2 =1.5
n1 =1
n = 1.6
r1, r2
Parallel light
plane for burning
Thin lens
a
V2
b
a. Assume that the front surface of the medium of n = 1.6 is at distance a = 10cm
and the plane in the medium is at distance b = 3cm. Determine the radii of curvature
of the lens, and find the focal length.
b. Assume that the radii of curvature of the lens are r1 = 11cm and r2 = -11cm and
determine a for b = 3.5cm.
PG26.Corner Mirror.(p.71)
Show that a corner mirror will reflect incident light
through an angle of 180o regardless of the angle
the incident light makes with the mirror.
PG27.Reflectivity and Eigenvalues.(see p.76)
Go to FileFig.(G33RESCY) and assume that the radii r1 and r2 are both 1 but that we
have a misalignment in distance d. Make a plot of the eigenvalues 1 and 2 for d =
1.9, 1.99 and so on to 1.99999. At what distance is the misalignment tolerable?
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