qN ho /N

/ ho /p

M = /

= ho/pho/N

= Np

qfN

pNM

111

The closest image that you could clearly view:

111

f

N

NfNM

Angular magnification

The farthest an image can be is?

f

N

fN

fNM

0111

A jeweler whose near point is 40 cm, examines a diamond with a small magnifying glass. The lens has a focal length of 5 cm and the image is -185 cm from the lens.

a)Determine the angular magnification.

b)Where should the image be located for viewing without eyestrain? What is the magnification then?

c)What maximum magnification is possible?

22.8cm0.40)cm185

1

cm5

1()

11( N

qfp

NM

cmf

NM cm

cm 00.850.40 Want q= , so 1/q = 0

cmf

NM cm

cm 00.911 50.40 When q=N

Is the eye depicted above nearsighted or farsighted?

1) nearsighted

2) farsighted

3) could be either

1 f

1 p

1 q

= +

An eyeglass prescription calls for lens powers of +2.1 D for the left eye and +2.5 D for the right eye. The positive power means the lens is converging. This person must be

A) nearsighted

B) farsighted

C) cannot tell from information given

p = 12 cm f1 = 8 cm

40 cm

p2 = 40 cm – q1 = 16 cm f2 = 12 cm

2

1

cm 16

1

cm 12

1

q

cm 48

341

2

q

cm 482 q 3cm 16

cm 482 M

Note: overall M = M1M2 = 6

q

1

cm 12

1

cm 8

1

cm 24

231 q

cm 24q2

cm 12

cm 24 M

p = 12 cm f1 = 8 cm

32 cm

p2 = 32 cm – 24 cm f2 = 12 cm

Move the lensescloser togetherso that the 1st

image is within the 2nd’s focal point.

2

1

cm 8

1

cm 12

1

q

cm 24

321

2

q

cm 242 q 3cm 8

cm 24-2 M

Note: overall M = M1M2 = 6

p = 12 cm f1 = 8 cm

16 cm

p2 = 16 cm – 24 cm f2 = 12 cm

What if you move the lenses

so close togetherthe 2nd interrupts

the formation of the 1st image?

2

1

cm 8

1

cm 12

1

q

The object

is virtual! cm 24

321

2

q

cm 8.42 q 6.0cm 8-

cm 4.82 M

Note: overall M = M1M2 = 1.2

Is the eye depicted above nearsighted or farsighted?

A) nearsighted

As an object is moved closer to the eye, the convergence point of the rays moves back (toward the retina). So this eye can focus on objects that are closer than the “normal” eye can focus.

B) farsighted

A positive lens power means the lens is convex, or converging. So without the lens, the rays must converge behind the retina, and so the

person is farsighted.