English Homework for Chapter 1
1.In 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 he ld erect casts a shadow 3.4m long, while a building’s shadow is 170m long. How tall is the building?
Solution. According to the law of rectilinear propagation, we get, 4.32170=
x x=100 (m)
So the building is 100m tall.
2.Light from a water medium with n=1.33 is incident upon a water-glass interface at an angle of 45o. The glass index is 1.50. What angle does the light make with the normal in the glass? Solution. According to the law of refraction, We get,
'
'sin sin I n I n =
626968
.05.145sin 33.1sin =⨯='
I
8.38='I
So the light make 38.8o with the normal in the glass.
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.
r n n l n l n -'=-'' and n ’=1 , n=1.33, r=-20
we can get
11416.110133
.15836.8)(5836.81165.02033.01033.11>-=⨯⨯-=''=
-='∴-=--+-=-'+='l n l n cm l r n n l n l β
So the fish appears larger.
A
4.An 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=1.
5. Find the image distance. Solution. Refer to the figure. According to the equation
r n n l n l n -'=-'' and n=1, n ’=1.5, l 1=-2cm, r 1=1cm , we get
cm l l d l l l 2021115.15.12
121
1='∴-∞='-=∞='∴=-+-='
English Homework for Chapter 2
1.An 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 l l '=-'11 and l=-30cm f ’=10 cm.
we get
)(15)30(10)
30(10cm l f l f l =-+-⨯=+''=
'
Others are omitted.
2.A 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 l l '=-'11 and f′=30cm l =-50cm
we get
)(75)50(30)
50(30cm l f l f l =-+-⨯=+''=
'
5.15075-=-='=
l l β
The image is a real, larger one.
3.The 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,
)(3020)60()
20()60(111
cm f l f l l +=+-⨯-='+'='
For the front surface (the face farther away from the lens),
)
(9.29204.6020
)4.60(2
cm l +=+-⨯-='
The transverse magnification for the rear surface is
⨯
-=-+=
5.06030
t M
But the axial magnification is
⨯+=----=∆'∆=
25.0)4.60(609
.2930l l M a
Since a t M M ≠,the cube doesn’t look like a cube.
′
