4.Example: Multiple mirrors and virtual images: how far away are the virtual images? 1m3m VII....

4. Example: Multiple mirrors and virtual images: how far away are the virtual images?

1m3m

VII. MirrorsVII. Mirrors


2m3m


6m 2m


6m 2m 8m


2m 8m 10m 16m

18m

d) Example: A telescope is designed to take images of distant galaxies. If the

diameter of the telescope’s spherical mirror is 40 cm, where should a detector be placed?

*Step 1: Stars ~ infinitely far away*Step 2: Use mirror equation:

1/s + 1/s’ = 2/R => 0 + 1/f = 2/R, so

f = R/2 = 10 cm.

VII. MirrorsVII. Mirrors

A. Refraction at a spherical surface 1. Images

a) Object distance, Image distance, and Radius of curvature

s

n1 < n2

s’

1

2R

r

Exterior angle of a triangle:1 = =

VIII. LensesVIII. Lenses



s

n1 < n2

s’

1

2R

r

Snell’s Law:n1sin1 =n2sin




s

n1 < n2

s’

1

2R

r

Tangents:tan= r/(s + ).tan= r/(s’ - ).tan = r/(R - ).




s

n1 < n2

s’

1

2R

r

Small angle approximation:tan~ ( << 1).sin~




s

n1 < n2

s’

1

2R

r

•Rewrite Snell’s Law:

n11 ~ n22, so

2 = (n1/n2)().




s

n1 < n2

s’

1

2R

r

* Use exterior law:

(n1 n2) = (n2 - n1)




s

n1 < n2

s’

1

2R

r* Small angles:

tan ~ ~ r/s ( << s).tan ~ ~ r/s’ (<<s’).tan ~ ~ r/R ( << R).


*Put it all together!

n1/s + n2/s’ = (n2 - n1)/R. (VIII.A.2)



s

n1 < n2

s’

1

2R

r



b) Magnification

s

n1 < n2

s’

1

2

h’

tan1 = h/s.tan2 = -h’/s’.

M = h’/h = -s’tan2/(stan1).

h


* Use Snell’s Law & small angle

M = h’/h ~ -s’2/(s1) = -n1s’/(n2s). (VIII.A.3)


b) Magnification

s

n1 < n2

s’

1

2

h’h


3. Example: An insect is centrally embedded in a spherical globule of amber. The index of refraction for the amber is 2 and the diameter of the globule is 4 cm. How far from the surface does the insect appear?

n1/s + n2/s’ = (n2 -n1)/R;2/(2 cm) + (1/s’) = (-1)1/(2 cm);1/s’ = 1/(4 cm) - 1/(1 cm) = -3/4 cm-1.

s’ = -4/3 = -1.333 cm: virtual image


Magnification:M = -n1s/n2s’ = -2*(-4/3)/1*2 = 4/3

B. Refraction at a flat surface1. Let R => infinity. Sphere => plane

n1/s = -n2/s’, or s’ = -(n2 /n1)s. (VIII.B.1)

n1 > n2.


B. Refraction at a flat surface2. Example: A frog sits at the bottom of a

reflecting pool. The pool is .5 m deep with n = 4/3. Where does the frog appear?

s’ = -(n2 /n1)s = -(1/[4/3])(0.5 m) = 3/8 m.

n1 > n2.


C. Thin Lenses2. Ray tracing: single lens


C. Thin Lenses2. Ray tracing: double lens


C. Thin Lenses6. Example: What is the image distance for a

thin glass lens with a focal length of 4/3 cm and an object at 10 cm?

1/s’ =1/f - 1/s = 1/(4/3 cm) - 1/10cm = .65 cm-1.Thus, s’ = +1.5 cm.

How is the image magnified? M = -s’/s = -1.5/10;M = -.15


7. Example: What is the image distance and magnification for the following pair of lenses:Lens 1: R1 = infinity, R2 = -10 cm, n = 1.5

Lens 2: R1 = 10 cm, R2 = -10 cm, n = 1.5The lenses are separated by 50 cm, and the object is at a distance of 45 cm from the first lens.

Step 1: Find the focal lengths1/f1 = (1/2)(0 - 1/(-10cm) => f1 = 20 cm.1/f2 = (1/2)(1/10cm - 1/(-10cm)) => f2 = 10 cm.




Step 2: Find image distance for lens 1:1/s’1 = 1/f1 - 1/s1 = 1/20cm - 1/45cm => s’1 = 36 cm.




Step 3: Find image magnification for lens 1:M1 = -s’1 = -36 cm/45 cm = -0.8



Lens 2: R1 = 10 cm, R2 = -10 cm, n = 1.5

Step 4: Find image distance for lens 2 using image 1 as object:

1/s’2 = 1/10 cm - 1/14 cm => s’2 = 35 cm.

The image is to the right of the second lens.



Lens 2: R1 = 10 cm, R2 = -10 cm, n = 1.5

Step 5: Find total magnificationM2 = -s’2/s2 = -35 cm/14 cm = -2.5.

Total magnification: M = M1* M2 = (-0.8)*(-2.5) = 2


4.Example: Multiple mirrors and virtual images: how far away are the virtual images? 1m3m VII....

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