Physics 150 Op-cal instruments Chapter 24 Real image q>0 Real object p>0 Virtual image q<0 Virtual object p<0 Upright image h'>0 Inverted image h'<0 Converging mirror (concave) f>0, f = ½ R Diverging mirror (convex) f<0, f = -­‐½ R Converging lens (convex) f>0 Diverging lens (concave) f<0 Physics 150, Prof. M. Nikolic 2 Lenses in combina-on Suppose we have more than one lens. How do we find the image of the combina-on? With lenses in combina-on, the image formed by one lens is the object for the next lens. Lens equa-on: 1 1 1
+ =
p1 q1 f1
1 1 1
+ =
p2 q2 f2
p2 = s − q1
Physics 150, Prof. M. Nikolic p is the object distance q is the image distance f is the focal length s is the distance between the lenses 3 Transverse magnifica-on The transverse magnifica-on of a combina-on of lenses is just the product of the magnifica-ons for the individual lenses. mtotal = m1m2 !mn
It is possible to have the object for the second lens be virtual (p2<0). Here the image formed by the first lens is beyond the second lens. Physics 150, Prof. M. Nikolic 4 Exercise: Lenses in combina-on A converging lens with a focal length of 15 cm and a diverging lens are placed 25 cm apart, with the converging lens on the le`. An object is placed 20 cm to the le` of the converging lens. The final image is upright and 35 cm to the le` of the converging lens. a)  What is the focal length of the diverging lens? b)  What is the total magnifica-on? p1 What is given: f1 = 15 cm s = 25 cm p1 = 20 cm The image is virtual: q2 = -­‐ 35 cm – 25 cm = -­‐ 60 cm s f1 q2 1 1 1
+ =
p1 q1 f1
Physics 150, Prof. M. Nikolic 1 1 1
+ =
p2 q2 f2
p2 = s − q1
5 Exercise: Lenses in combina-on A converging lens with a focal length of 15 cm and a diverging lens are placed 25 cm apart, with the converging lens on the le`. An object is placed 20 cm to the le` of the converging lens. The final image is 35 cm to the le` of the converging lens. a)  What is the focal length of the diverging lens? b)  What is the total magnifica-on? What is given: f1 = 15 cm s = 25 cm p1 = 20 cm q2 = -­‐ 35 cm – 25 cm = -­‐ 55 cm Start with the first lens equa-on to find the posi-on of the image due to the first object: 1 1 1
+ =
p1 q1 f1
1 1 1
= −
q1 f1 p1
1
1
1
4−3 1
=
−
=
=
q1 15cm 20cm
60
60
q1 = 60cm
Then use third equa-on to relate the image of the first object as the new object for the second lens: p = s − q
2
1
Physics 150, Prof. M. Nikolic p2 = 25cm − 60cm = −35cm
6 Exercise: Lenses in combina-on A converging lens with a focal length of 15 cm and a diverging lens are placed 25 cm apart, with the converging lens on the le`. An object is placed 20 cm to the le` of the converging lens. The final image is 35 cm to the le` of the converging lens. a)  What is the focal length of the diverging lens? b)  What is the total magnifica-on? What is given: f1 = 15 cm s = 25 cm p1 = 20 cm q2 = -­‐ 35 cm – 25 cm = -­‐ 60 cm Finally, use the second lens equa-on to find the focal point of the divergent lens: 1 1 1
+ =
p2 q2 f2
1
1
1
=
+
= −0.0286 − 0.0167 = 0.0452
f2 −35cm −60cm
f2 = 22.1cm
b) Total magnifica-on q1
q2
mtotal = m1m2 = − × −
p1
p2
Physics 150, Prof. M. Nikolic mtotal = −
60
(−60)
×−
= 5.14
20
(−35)
7 The camera One of the simplest op-cal instruments •  One converging lens (most o`en) •  Real, inverted image Loca-on of film Physics 150, Prof. M. Nikolic 8 The camera A camera adjusts the loca-on of the lens so that the image distance just happens to be the distance from the lens to the film. For far away objects, the film must be placed one focal length from the lens or the image distance is equal to the focal length. Physics 150, Prof. M. Nikolic 9 Exercise: Cameras A person on safari wants to take a photograph of a hippopotamus from a distance of 75.0 m. The animal is 4.00 m long and its image is to be 1.20 cm long on the film. (a)  What focal length lens should be used? (b)  What would be the size of the image if a lens of 50.0 mm focal length were used? What is given: p = 75 m object width = 4 m width on film = -­‐1.2 cm (inverted image) width on film h!
q
m=
= =−
object width h
p
width on film
q
=−
object width
p
width on film
q=−
×p
object width
" −1.2 cm %
q = −$
' ( 75.0 m ) = 22.5 cm
# 4.0 m &
Physics 150, Prof. M. Nikolic Posi-ve sign – real image 10 Exercise: Cameras A person on safari wants to take a photograph of a hippopotamus from a distance of 75.0 m. The animal is 4.00 m long and its image is to be 1.20 cm long on the film. (a)  What focal length lens should be used? What is given: p = 75 m object width = 4 m width on film = -­‐1.2 cm (inverted image) Then we use the thin lens equa-on 1 1 1
+ =
p q f
1
1
1
=
+
= 4.458 1
m
f 75m 0.225m
f = 22.4 cm
Physics 150, Prof. M. Nikolic Focal length is very similar to the image distance è true for far away objects 11 Exercise: Cameras A person on safari wants to take a photograph of a hippopotamus from a distance of 75.0 m. The animal is 4.00 m long and its image is to be 1.20 cm long on the film. (b) What would be the size of the image if a lens of 50.0 mm focal length were used? 1 1 1
1 1 1
+ = ⇒ = −
p q f
q f p
1
1
1
1
=
−
≈ 20
q 0.05cm 7500cm
cm
q = 50.0 mm
m=
width on film h!
q
= =−
object width h
p
q
width on film = − × object width
p
" 50.0 mm %
width on film = − $
' ( 4.0 m ) = −2.67 mm
# 75.0 m &
The image would be 2.67 mm long, real and inverted. Physics 150, Prof. M. Nikolic 12 The eye The human eye is very similar to digital camera •  Light passes through the converging lens of the eye •  And is projected (upside down) on the light-­‐sensi-ve cells of the re,na •  Rods •  Cones The eye lens is at a fixed distance from the re-na (unlike the camera) but it can adjust its focal length, thanks to the ciliary muscle. Addi-onal focusing is done curving the cornea. Pupils controlled by the iris provide an adjustable aperture. Physics 150, Prof. M. Nikolic 13 The eye The near point is the closest distance from your eye that an object can be seen clearly. For a normal eye this distance is 25 cm. The far point is farthest distance from your eye that an object can be seen clearly. For a normal eye this distance can be arbitrarily far away (infinite or ∞). Blind spot is a place located on re-na where op-c nerve leaves the re-na with no rods and cones When an optometrist prescribes glasses, they usually do it in terms of the “power” of the lens. The refrac,ve power of a lens is defined to be: 1
P=
f
Physics 150, Prof. M. Nikolic where f is the focal length of the lens; typical units of P are diopters (1D = 1 m-­‐1). 14 Correc-ng myopia For a near sighted (myopic) person, light rays of the distant objects converge before they strike the re-na. A diverging lens is placed in the light path. This creates a virtual image closer to your eye than the actual object is (at the posi-on where the eye can focus it). Focal length and refrac-ve power of the lens are nega-ve. Physics 150, Prof. M. Nikolic 15 Correc-ng hyperopia For a far sighted (hyperopic) person, light rays of the nearby objects converge a`er they strike the re-na. A converging lens is placed in the light path. This creates a virtual image farther from your eye than the actual object is (at the posi-on where the eye can focus it). Focal length and refrac-ve power of the lens are posi-ve. Physics 150, Prof. M. Nikolic 16 Exercise: Correc-on for farsighted eye If Harry has a near point of 1.5 m, what focal length contact lenses does he require? The near point refers to the closest distance an object can be to see it clearly, in this case 1.5 m. A normal eye has a near point of p = 25 cm. These correc-ve lenses must take an object at p=25 cm and form a virtual image at a distance of q=-­‐1.5 m. We should use the thin lens equa-on: 1 1 1
+ =
p q f
1
1
1
1
=
+
= 3.33
f 0.25m −1.5m
m
f = 0.3m
The refrac-ve power of these lenses is P=
1
f
Physics 150, Prof. M. Nikolic 1
P=
= +3.3 D
0.30 m
Farsighted people use converging lenses with posi-ve focal length and refrac-ve power. 17 Exercise: Correc-on for nearsighted eye A person comes in to the doctors office with a far point of 1.75 m. What focal length contact lenses does he require? The far point refers to the farthest distance an object can be to see it clearly, in this case 1.75 m. A normal eye has a far point of p = ∞. These correc-ve lenses must take an object at p=∞ and form a virtual image at a distance of q=-­‐1.75 m. We should use the thin lens equa-on: 1 1 1
+ =
p q f
1 1
1
1
= +
= 0 − 0.57
f ∞ −1.75m
m
f = −1.75m
The refrac-ve power of these lenses is P=
1
f
Physics 150, Prof. M. Nikolic 1
P=
= -0.57 D
1.75 m
Nearsighted people use diverging lenses with nega-ve focal length and refrac-ve power. 18 Angular magnifica-on The farther an object is from your eye, the smaller it will look. For an object to look bigger the image of it formed on the re-na must be made bigger. θaided
θunaided
θ
θ is the angle between the rays from top and the bokom of each object and is called the angular size
θ aided
The angular magnifica-on is M =
θ unaided
Physics 150, Prof. M. Nikolic 19 A simple magnifier A simple magnifier is a converging lens placed so that the object distance is less than the focal length (it produces enlarged and upright virtual image). N The closest this object can be held and s-ll be seen clearly is at the near point, N=25 cm. h tan θ unaided ≈ θ unaided =
h
N
f When an object is placed close the focal point, the extrapolated lines turn out to be almost parallel. The image is highly magnified and very distant. The eye can relax and observe that image with no strain tan θ aided ≈ θ aided =
Physics 150, Prof. M. Nikolic h
p
20 A simple magnifier The angular magnifica-on h
θ aided
p N
M=
=
=
h
θ unaided
p
N
N is the near point for a person (typically 25 cm) p is the distance of the object from the magnifier What happens if you place an object at the focal point? Virtual image will be formed at infinity. p= f
M=
θ aided N
=
θ unaided f
f is the focal length of the lens used in the magnifier Physics 150, Prof. M. Nikolic 21 Exercise: Magnifying glass An insect that is 5.00 mm long is placed 10.0 cm from a magnifying glass (converging lens) with a focal length of 12.0 cm. (a) What is the size of the image? What’s given h = 5 mm p = 10 cm f = 12 cm We could use the equa-on for the transverse magnifica-on h!
q
m= =−
h
p
1 1 1
+ =
p q f
But then we have to find the distance of the image from the magnifying glass. 1 1 1
= −
q f p
1
1
1
1
=
−
= −0.0167
q 12cm 10cm
cm
q = −60cm
Go back to the first equa-on and solve for h’ q
h! = − h
p
Physics 150, Prof. M. Nikolic −60cm
h! = −
5mm = 30mm
10cm
22 Exercise: Magnifying glass An insect that is 5.00 mm long is placed 10.0 cm from a magnifying glass (converging lens) with a focal length of 12.0 cm. What’s given h = 5 mm p = 10 cm f = 12 cm And we found: q = -­‐60 cm h’ = 30 mm (b) Is the image upright or inverted? Since h’>0, the image is upright (c) Is the image real or virtual? Since q < 0 (the image forms on the same side of the lens as the object), the image is virtual. (d) What is the angular magnifica-on if the lens is close to the eye? θ aided N
M=
=
θ unaided p
Physics 150, Prof. M. Nikolic 25 cm
M=
= 2.5
10 cm
23 The compound microscope Two converging lenses are used to produce a highly magnified image. f0
fe
f0
•  The object is placed outside of the focal point of lens 1 called objec,ve and it forms an enlarged, real image just inside the focal point of the lens 2 called eyepiece •  The eyepiece is used to form and view the distant, large, and virtual image – just like a simple magnifier. Physics 150, Prof. M. Nikolic 24 The compound microscope The total magnifica-on is the product of the individual magnifica-ons of the two lenses: " L %" N %
M total = mobjmeye = − $ '$ '
# fo & # fe &
L is the tube length, distance between focal points: L = q0 − f0
N = 25 cm is near point distance fo is the focal length of the objec-ve fe is focal lengths of the eyepiece And the thin lens equa-ons s-ll apply 1 1 1
+ =
p0 q0 f0
Physics 150, Prof. M. Nikolic 1 1 1
+ =
pe qe fe
25 Exercise: Op-cal microscope A microscope has an objec-ve lens of focal length 5.00 mm. The objec-ve forms an image 16.5 cm from the lens. The focal length of the eyepiece is 2.80 cm. Objec-ve image is formed at the focal point of the eyepiece. (a) What is the distance between the lenses? q0
f0
fe
What’s given: f0 = 5 mm q0 = 16.5 cm fe = 2.8 cm f0
Distance between the lenses is: d = q0 + fe = 16.5 cm + 2.8 cm = 19.3 cm Physics 150, Prof. M. Nikolic 26 Exercise: Op-cal microscope A microscope has an objec-ve lens of focal length 5.00 mm. The objec-ve forms an image 16.5 cm from the lens. The focal length of the eyepiece is 2.80 cm. Objec-ve image is formed at the focal point of the eyepiece. (b) What is the total angular magnifica-on? The near point is 25.0 cm. What’s given: f0 = 5 mm q0 = 16.5 cm fe = 2.8 cm " L %" N %
M total = mobjmeye = − $ '$ '
# fo & # fe &
And we know that the tube length is: L = qo – fo L = 16.5 cm – 0.5 cm = 16 cm " 16cm %" 25cm %
M total = − $
'$
' = −285.7
# 0.5cm &# 2.8cm &
(c) How far from the objec-ve should the object be placed? 1 1 1
+ =
p0 q0 f0
Physics 150, Prof. M. Nikolic 1
1 1
= −
p0 f0 q0
1
1
1
1
=
−
= 1.94
p0 0.5cm 16.5cm
cm
p0 = 0.52cm
27 The telescope A telescope is a combina-on of lenses and/or mirrors used to collect a large amount of light and bring it to a focus. The most common type of telescope is the refrac,ng telescope •  Light rays from distant object are almost parallel to the first lens (objec-ve) è produces small, real and inverted image at its focal point •  This image is placed just outside the focal point of the second lens (eyepiece, ocular) è produces distant, enlarged and virtual final image Physics 150, Prof. M. Nikolic 28 Refrac-ng telescope The angular magnifica-on is: θe
fo
M = =−
θo
fe
Physics 150, Prof. M. Nikolic The barrel (or tube) length is: L = fo + fe
29 Reflec-ng telescope A reflec-ng telescope uses combina-on of the mirrors and lenses. Applica-on – Hubble space telescope (Cassagrain telescope) Physics 150, Prof. M. Nikolic 30 Exercise: Telescopes A refrac-ng telescope is used to view the moon (diameter of 3474 km & distance from Earth 384,500 km). The focal lengths of the objec-ve and eyepiece are +2.40 m and +16.0 cm, respec-vely. (a) What should be the distance between the lenses? What’s given: Object diameter: h = 3474 km p = 384,500 km fo = +2.4 m fe = +16 cm L = fo + fe
L = 2.4m + 0.16m = 2.56m
(b) What is the angular magnifica-on? θe
fo
M = =−
θo
fe
Physics 150, Prof. M. Nikolic 2.40 m
M =−
= −15
0.16 m
31 Exercise: Telescopes A refrac-ng telescope is used to view the moon (diameter of 3474 km & distance from Earth 384,500 km). The focal lengths of the objec-ve and eyepiece are +2.40 m and +16.0 cm, respec-vely. (c) What is the diameter of the image produced by the objec-ve? What’s given: Object diameter: h = 3474 km p = 384,500 km fo = +2.4 m fe = +16 cm image size h!
q
m=
= =−
object size h
p
q
h! = − h
p
The image distance q, is equal to the focal length of of the objec-ve lens: q = 2.4 m # 2.40 m &
h! = − %
( (3474 m ) = 2.17 cm
$ 384,500 km '
Physics 150, Prof. M. Nikolic 32