工程光学英文题加中文题含答案

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 held 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, 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

4.32

170

x

refraction, We get,

So the light make

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. and n ’

=1 , n=1.33, r=-20

we can get

So the fish appears larger.

'

'

sin sin I n I n =626968

.05

.145

sin 33.1sin =⨯=

'

I

8

.38='I r

n

n l

n

l n -'=

-

''

1

1416.110

133

.15836.8)(5836.81165

.020

33.010

33.11>-=⨯⨯-=

''=

-='∴-=--+-=-'+='l

n l n cm l r n n l n l β n 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

and n=1, n ’

=1.5, l 1=-2cm,

r

n n l n l n -'=-''

r 1=1cm , we get

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? V erify your answer by graphical construction of the image. Solution. According to

equation, and l=-30cm f ’

we get

Others are omitted.

cm l l d l l l 20

2

11

15.15.12

1211='∴-∞='-=∞='∴=-+-='

f l l '

=

-'

1

1

)

(15)

30(10)

30(10cm l

f l

f l =-+-⨯=

+''=

'′

′

′

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.

and f′=30cm l we get

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,

f l l '

=

-'

11(75)

50(30)50(30l

f l f l =-+-⨯=

+''=

'5

.150

75-=-='=

l l β)

(3020

)60()20()60(111

cm f l f l l +=+-⨯-=

'

+'='′