F1
Figure 27a (in Answer T5) shows a diagram similar to that
required, but with different dimensions. The object is between the
first focus and the lens. The image is erect and virtual. The lateral
magnification m = v/u :
m = (−25 cm)/(–10 cm) = +2.5. (v is the image distance, u is the
object distance and the Cartesian sign convention has been used.)
1
F2
1
0
5
10 cm
(a)
Figure 27a See Answer T5. A convex lens
used as a magnifying glass.
3
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
F2
Use the thin lens equation (Equation 12):
1 1
1
− =
v u
f
The thin lens equation (Cartesian sign convention)
(Eqn 12)
1/v – 1/u = 1/f, with u = –25 cm, f = +10 cm, so that:
1
1
1/v = (1/10 cm) + (–1/25 cm) = (5−2)/(50 cm) = 3/(50 cm) and v = +50 cm/3 = +16.7 cm.
1
1
1
1
1
1
The image is real and inverted with a magnification m = v/u = 16.7 cm/25 cm = 0.67.
1
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
1
B
object
X
0
h
C
5 cm
F2
I
F1
O
h'
image
M
Y
v
u
Figure 19 Ray diagram for a convex lens with f = 10 cm and u = –15 cm. Note that the zero on the scale bar does not
correspond to the origin of Cartesian coordinates which is at C.
3
1
1
Your sketch should resemble Figure 19 but with the object further from the lens, and the image closer and
smaller than the object.
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
F3
The ray diagram is similar to Figure 24.
Use the formula 1/u + 1/v = 1/f with
u = −15 cm and f = −10 cm.
1
(1)
B
1
Then 1/v = (1/−10 cm) − (1/−15 cm)
= (−3 + 2)/(30 cm)= −1/(30 cm), and
v = −30 cm.
1
1
1
P
β
β
(2)
h
β
I
R
1
2β
O
α
α
F
C
1
(1)
The image is real, inverted and 30 cm in front
of the mirror.
1
(2)
M
f
u
r
v
Figure 24 A spherical concave mirror forming a real inverted
image of an extended object.
3
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route
through the module and to proceed directly to Ready to study? in Subsection 1.3.
Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the
Closing items.
If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it
here.
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
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S570 V1.1
R1
(i) The incident ray, the reflected ray and the normal at the reflection point all lie in the same plane.
(ii) The angle of the incident ray to the normal equals the angle of the reflected ray to the normal.
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
R2
Snell’s law says: µair1sin1θ air = µglass1sin1θglass. The ray is refracted towards the normal since µglass > µair ≈ 1.00.
We have:
θ glass = arcsin ( sin θ air µ glass ) = arcsin ( 0. 259 1. 50 )
= arcsin ( 0.173 )
θ glass = 9. 9°
The refractive index µglass = c/v, where v is the speed of light in the material and c is the speed of light in a
vacuum.
v = 3.0 × 108 m s−1/1.5 = 2.0 × 108 m s−1
1
1
1
1
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
R3
At 10°, tan1θ = 0.1763 and θ = 10π/180 = 0.17451radians.
The percentage error is: (0.0018/0.1763) × 100% = 1%.
At 10°, sinθ = 0.1736 and the percentage error is: (0.0009/0.1736) × 100% = 0.5%.
The approximation improves as θ gets smaller.
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
T1
The apex angle A of the prism is 60°. Using Equation 5 for Dmin:
Dmin = 2 arcsin0[µ sin(A/2)] – A
blue light:111111100 0Dmin = 2 × 49.42° − 60° = 38.84°
green light:111100 0 Dmin = 2 × 49.24° − 60° = 38.49°
yellow light: 0 0 Dmin = 2 × 48.94° − 60° = 37.88°
red light:11111111100 0 0 Dmin = 2 × 48.68° − 60° = 37.35°
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
(Eqn 5)
T2
In the paraxial approximation, µθ = µ′θ ′
From the triangles of Figure 10 we have θ = β – α and
θ0′ = β – γ , so that µ(β – α) = µ′ (β – γ)
Replacing angles by their tangents then gives
β = tan1 β = h/r,
α = tan1 α = h/l, γ = tan1 γ = h/l′
so
h h
h h
µ  −  = µ ′ − 
r
 r l′ 
l
and
−
µ µ′
µ′ − µ
+
=
l′
l
r
This differs from Equation 7
µ µ′
µ′ − µ
+
=
l′
l
r
in the sign of the first term.
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
(Eqn 7)
θ'
µ
γ
α
O
µ'
θ
β
C
R
O'
r
l'
l
Figure 10
The point image of a point object formed by a concave boundary surface (see Question T2).
3
T3
Referring to Figure 10 we see that O, O′ and R all lie to the left of the origin at C so that l, l′ and r must all be
made negative. With these changes, the equation which forms the answer to Question T2 is unchanged.
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
T4
Use the lens maker’s equation:
1/f = (µ − 1)(1/r1 − 1/r2 )
(Eqn 11)
(1/15)]1 cm−1
Lens A: 1/f = 0.5[(1/25) −
= [0.5(3 −
= (−1/75) cm−1.
So f = −75 cm. The lens is diverging with convex front face and concave back face.
(See Figure 26a.)
Lens B: 1/f = 0.5[(1/−10) – (1/–5)] cm−1 = [0.5(−1 + 2)/10] cm−1 = (1/20) cm−1.
So f = 20 cm. The lens is converging with concave front face and convex back face.
(See Figure 26b.)
5)/75]1 cm−1
1
1
1
1
(a)
(b)
1
1
Lens C: 1/f = 0.5[(1/±∞) – (1/–20)] cm−1 = 0.5(0 + 1/20) cm−1 = (1/40)1cm−1.
So f = 40 cm. The lens is planoconvex and is converging. Note that the ‘radius of
curvature’ of a plane surface may be taken as ±∞; it makes no difference.
(See Figure 26c.)
Lens D: 1/f = 0.5[(1/20) – (1/20)] cm−1 = 0 cm−1. So f = ∞. The front surface is convex
and the rear surface is concave with equal curvature. There is zero net refraction and rays
pass through with no change of direction. (See Figure 26d.)
1
1
1
1
1
(d)
(c)
Figure 26 See
Answer T4.
3
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
T5
(a) This is an example of a convex lens used as a magnifying
glass. Using Equation 12
1 1
1
The thin lens equation
− =
(Eqn 12)
v u
f
with f = 10 cm and u = –6 cm gives
1/v = (1/10 cm) + (1/–6 cm) = (3 – 5)/(30 cm) = –2/(30 cm)
so that v = –15 cm.
The ray diagram is shown in Figure 27a.
1
1
1
1
1
F2
1
1
0
5
10 cm
(a)
Figure 27a
See Answer T5. A convex lens
3
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
(b) This is more difficult to visualize, but Equation 12
1 1
1
The thin lens equation
− =
(Eqn 12)
v u
f
takes care of the calculation without any difficulties.
M
C
I
real
image
O
virtual
object
F2
Q
As the object is virtual, it must be to the right of the lens, so that
u = +16 cm. The focal length f is +20 cm and the equation gives
1/v = (1/20 cm) + (1/16 cm) = (4 + 5)/(80 cm) = 9/(80 cm),
so that v = 8.9 cm.
1
B
X
P
0
(b)
5
10 cm
1
1
1
1
1
1
Figure 27b See Answer T5. A convex lens
produces a real image of a virtual object.
3
To construct the diagram we must choose two rays which would have met at the tip B of the arrow if the lens
were not there. Suitable rays are PX and QC in Figure 27b. QC proceeds undeviated to B but PX, being parallel
to the optical axis is refracted through F2 . The image of the arrow tip is formed at the intersection M of XF2 and
CB.
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
T6
It is a test of the care you have taken with your ray diagrams to see how closely your measured results come to
the calculated ones. The precise results obtained from calculation are as follows:
For Figure 19, u = –15 cm, v = 30 cm, m = 30 cm/(–15 cm) = –2, and the minus indicates an inverted image
For Figure 20, u = –15 cm, v = –6 cm, m = –6 cm/(–15 cm) = 0.4 and the image is erect.
For Figure 21, u = 80 cm, v = –133 cm, m = –133 cm/80 cm = –1.66 with an inverted image.
1
1
1
1
1
1
1
1
1
1
1
1
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
T7
Refer to Figure
28. It is necessary
to draw two sets
of principal rays.
The first set is
used to locate the
first image and
the second set to
locate the final
image produced
by the second
lens with the first
image as the
second object.
M1
B
I2
O
I1
M2
10 cm
Figure 28
See Answer T7.
3
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
For the first lens:
1/v1 = 1/f1 + 1/u1 = (1/10 cm) + (1/–51cm )= − 1/(10 cm)
1
1
so v 1 = −10 cm
1
The first image is virtual and erect. The distance of the second object (i.e. the first image) to the second lens is:
u2 = –(d − v 1 ) = –[35 − (−10)] cm = – 45 cm
1
1
1/v2 = 1/f2 + 1/u2 = (1/10 cm) + (1/−0451cm)
1
11
1= (9 − 2)/(90 cm) = 7/(90 cm) so v2 = 12.9 cm
1
1
1
The final image is real, inverted and to the right of the second lens. The overall magnification is given by the
product of the separate magnifications of the two lenses:
m = m1m2 = (v1 /u1 )(v2 /u2 )
1111001= [(–10 cm)/(–5 cm)][12.9 cm/(–45 cm)] = −00.57
1
1
1
1
The final real image is reduced in size by a factor 0.57 and it is inverted as the minus sign indicates.
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
T8
Use Equation 19:
1 1
1
− = −
u v
f
(Eqn 19)
with f = 20 cm and v = 15 cm. Then 1/u = (–1/20 cm) + (1/15 cm) = (–3 + 4)/(60 cm) = 1/(60 cm), and u = 60 cm.
Remember, this result made no use of the Cartesian sign convention.
1
1
1
1
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
1
1
1
T9
Use Equation 21a:
1 1
1
+ =
v u
f
(Eqn 21a)
with f = –20 cm.
1
(a) u = –30 cm so 1/v = (1/–20 cm) − (1/–30 cm)
= −1/(60 cm) and v = −60 cm
The image is in front of the mirror and therefore is real.
1
1
1
1
1
(b) Now we have u = −15 cm and the same focal length so 1/v = 1/(–20 cm) − 1/(–15 cm) = 1/(60 cm) to give
v = +60 cm so that the image is to the right of the mirror and is therefore virtual.
(c) With u = –0∞, 1/v = 1/(–20 cm) − 1/(–0∞) = –1/(20 cm) and v = –20 cm. The image is real and at the focus.
1
1
1
1
1
FLAP P6.3
Optical elements: prisms, lenses and spherical mirrors
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
1
1
1