English Homework for Chapter 1
ancient times the rectilinear propagation of light was used to measure the height of objects by comparing the length of their shadows with the length of the shadow of an object of known length. A staff 2m long when held erect casts a shadow long, while a building’s shadow is 170m long. How tall is the building
x2?Solution. According to the law of rectilinear propagation, we get, 1703.4
x=100 (m)
So the building is 100m tall.
from a water medium with n= is incident upon a water-glass interface at an angle of 45o. The glass index is . What angle does the light make with the normal in the glass
Solution. According to the law of refraction, We get,
'' nsinI?nsinI
1.33?sin45?sinI???0.6269681.5
n= 45 oI??38.8?
n′=So the light make with the normal in the glass.
I′
3. A goldfish swims 10cm from the side of a spherical bowl of water of radius 20cm. Where does the fish appear to be Does it appear larger or smaller Solution. According to the equation.
n?nn??n??A?l lr and n’=1 , n=, r=-20
we can get
1nn??n1.33?0.33??????0.1165l?lr?10?20?l???8.5836(cm)???nl??8.5836?1.33???1.1416?1n?l1?10
So the fish appears larger.
R1=20cm R2=-20cm A
object is located 2cm to the left of convex end of a glass rod which has a radius of curvature of 1cm. The index of refraction of the glass is n=. Find the image distance.
Solution. Refer to the figure. According to the equation
-10cm
n?nn??n???llr and n=1, n’=, l1=-2cm, r1=1cm , we get
1.51.5?11???0?l11?2????l1????l2?d?l1??2cm?l2
r1=1cm A -l1=2cm A′
l2′
English Homework for Chapter 2
object 1cm high is 30cm in front of a thin lens with a focal length of 10cm. Where is the image Verify your answer by graphical construction of the image. Solution. According to the Gauss’s equation,
f′=10cm 11?l?l?f? and l=-30cm f’=10 cm.
we get
y=1cm l??f?l10?(?30)??15(cm)f??l10?(?30)
-l=30cm Others are omitted.
lens is known to have a focal length of 30cm in air. An object is placed 50cm to the left of the lens. Locate the image and characterize it. Solution. According to Gauss’s equation,
f′=30cm
11?l?l?f? and f′=30cm l=-50cm
l??we get
f?l30?(?50)??75(cm)?f?l30?(?50)
-l=50cm
??l?75???1.5l?50
The image is a real, larger one.
object is transparent cube, 4mm across, placed 60cm in front of 20cm focal length.
Calculate the transverse and axial magnification and describe what the image looks like
Solution. From Gauss’s equation, we find for the rear surface of the cube (the face closer to the lens)
that,
??l1l1f?(?60)?(20)???30(cm)l1?f?(?60)?20
For the front surface (the face farther away from the lens),
??l2(?60.4)?20??29.9(cm)?60.4?20
Mt??30??0.5??60
The transverse magnification for the rear surface is
But the axial magnification is
Ma??l?30?29.9???0.25??l?60?(?60.4)
Since Mt?Ma,the cube doesn’t look like a cube.
biconvex lens is made out of glass of n=. If one surface has twice the radius of curvature of the other, and if the focal length is 5cm, what are the two radii Solution. Supposing r1= -2r2 (ρ2=-2ρ1),according to the lens equation
1??(n?1)(?1??2) we get, 5?(1.52?1)(?1??2)
r1 -r2 ??1?0.1282?2??0.2564
∴r1=(cm) r2=(cm)
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English Homework for Chapter 4
1. A stop 8mm in diameter is placed halfway between an extended object and a large-diameter lens of 9cm focal length. The lens projects an image of the object onto a screen 14cm away. What is the diameter of the exit pupil
Object Stop Lens Image -l l’ Solution. Refer to the figure. First, from the known focal length and the image distance,
we find the object distance.
?111??l?lf? and l’=14 f’=9 l=(cm)
The stop is one-half that distance is front of the lens, so ls=(cm) ∴ls=(cm)
’
???Dexl?31.5?s??25.2Dstopls2
∴
Dex?2.5?0.8?2(cm)2. Two lenses, a lens of focal length and a minus lens of unknown power, are mounted
工程光学练习题（英文题加中文题含答案）
