• Study Resource
• Explore

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
```Physics 112
Homework 12 (Ch24)
1. If an f  135 -mm telephoto lens is designed to cover object distances from 1.2 m to , over
what distance must the lens move relative to the plane of the film?
Solution
We locate the image from equation:
 1  1
 
 d o   di
 1
 .
 f
a) For an object at infinity, the image will be in the focal plane, so we have di  f  135mm.
b) When the object is at 1.2 m,
 1  1


 120 cm   di

1
, which gives di  152 mm.

 135cm
Thus the distance from the lens to the film must change by
dib  dia  152mm  135mm  17mm.
2. A person’s left eye is corrected by a  3.50-diopter lens, 2.0 cm from the eye. (a) Is this
person’s left eye near- or farsighted? (b) What is this eye’s far point without glasses?
Solution
(a) Since the diopter is negative, the lens is diverging, so it produces images closer than the
object; thus the person is nearsighted.
(b) We find the far point by finding the image distance for an object at infinity:
 1  1 1
       P;
 d o   di  f
1 1
       3.50D, which gives di   0.286 m   28.6cm.
    di 
Because this is the distance from the lens, the far point without glasses is
28.6cm  2.0cm  30.6cm.
Physics 112
Homework 12 (Ch24)
3. A 3.30-mm-wide beetle is viewed with a 9.50-cm-focal-length lens. A normal eye views the
image at its near point. Calculate (a) the angular magnification, (b) the width of the image,
and (c) the object distance from the lens.
Solution
(a) The angular magnification with the image at the near point is
M
 25.0cm   1  3.63  .
N
1 
f
 9.50cm 
hi
(b) Because the object without the lens and the image with the lens
are at the near point, the angular magnification is also the ratio of
widths:
' h
M   i  hi  Mh0 ;
 h0
hi  3.63  3.30mm ,
which gives hi  12.0mm.

ho
O
I
N
ho

O
(c) We find the object distance from
 1  1 1
    ;
 d o   di  f
 1  

1
1
cm,
 

 do    25.0cm  9.50
Method 2:
 ' 25cm

c) M  

d0
d0 
which gives d o  6.88 cm from the lens.
25cm 25cm

 6.88cm
M
3.632
4. A magnifying glass with a focal length of 8.5 cm is used to read print placed at a distance of
7.5 cm. Calculate (a) the position of the image; (b) the angular magnification.
Solution
(a) We find the image distance from
 1  1 1
    ;
 d o   di  f
 1  1 1
cm, which gives di   64cm.

  
 7.5cm   di  8.5
(b) The angular magnification is
  25cm  25cm 
=
 3.3  .
M 

do
 7.5cm 
Physics 112
Homework 12 (Ch24)
5. An astronomical telescope has its two lenses spaced 75.2 cm apart. If the objective lens has a
focal length of 74.5 cm, what is the magnification of this telescope? Assume a relaxed eye.
Solution
For both object and image far away, we find the focal length of the eyepiece from the separation
of the lenses:
L  fo  fe ;
f e  L  f 0  75.2cm-74.5cm = 0.7cm
The magnification of the telescope is given by
M 
 75.2cm   110  .
fo

fe
 0.7cm 
6. A microscope has a 1.8-cm-focal-length eyepiece and a 0.80-cm objective lens. Assuming a
relaxed normal eye, calculate (a) the position of the object if the distance between the lenses
is 16.0 cm, and (b) the total magnification.
Solution
(a) Because the image from the objective is at the focal point of the eyepiece, the image distance
for the objective is
dio  l  f e  16.0cm  1.8cm  14.2cm.
We find the object distance from the lens equation for the objective:
 1   1  1

   ;
 doo   dio  f o
 1   1 
1
, which gives doo  0.85cm.



 doo   14.2 cm  0.80cm
(b) With the final image at infinity (relaxed normal eye), the magnification of the eyepiece is
Me 
N  25.0cm 

 13.9  .
f
1.8cm 
The magnification of the objective is
d
14.2cm
mo  io 
 16.7  .
d oo 0.85cm
The total magnification is
M= mo M e  16.7 13.9  230 
Method 2:
 25cm  l  f e   25cm  16cm  1.8cm 

  
M  

  232
 f e  d0   1.8cm  0.85cm 
Physics 112
Homework 12 (Ch24)
7. Two stars 15 light-years away are barely resolved by a 55-cm (mirror diameter) telescope.
How far apart are the stars? Assume   550 nm and that the resolution is limited by
diffraction.
Solution
The resolution of the telescope is
9
1.22 1.22   550  10 m 

