## November 5, 1999

### Low: 46

Possibly useful information:

For reflection, inc = ref

For refraction, n1 sin 1 = n2 sin 2

n = c/v

image equation

apparent depth di = do (n1/n2)

Young's Double Slit: = d sin

d sin = m , maxima for m = 0, 1, 2, . . .

d sin = (m + 1/2) , minima for m = 0, 1, 2, . . .

diffraction grating: = d sin

d sin = m , maxima for m = 0, 1, 2, . . .

diffraction by a slit: = D sin

First minimum for = D sin =

sin = /D

/D

minima for = D sin = m , m = 1, 2, 3, . . .

optical resolution: 1.22 ( /D )

polarization: S = So cos2

thin films: ' = /n

2t = m ' = m(/n); max, m = 0, 1, 2, . . . n1 < n2 < n3

2t = (m+1/2) ' = (m + 1/2 )(/n);

min, m = 0, 1, 2, . . . n1 < n2 < n3

2t = m ' = m(/n); min, m = 0, 1, 2, . . . n1 < n2 > n3

2t = (m + 1/2) = (m + 1/2)(/n);

max, m = 0, 1, 2, . . . n1 < n2 > n33

Simple Magnifier: Mang = r / f = 25 cm / f

Microscope:

Telescope:

We have covered lots of good things in this section! There may, indeed, be too many questions. So budget your time wisely.

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1. (23.20) A SCUBA diver, underwater, shines a light up toward the smooth surface of the water (n = 1.33) with an angle of incidence of 30°. At what angle does the light leave the water?

We begin with Snell's law,

n1 sin 1 = n2 sin 2

(1.33)(sin 30°) = (1.00) sin 2

(1.33)(0.5) = (1.00) sin 2

sin 2 = 0.665

2 = 41.7°

2. (23.43) Through what angle should an analyzer be rotated from the incoming plane of polarization to reduce the intensity to one-half?

The intensity is given by

S = So cos2

[This is contained in the text, on page 885, before Example 23.8; the equation given in Example 23.8, at the top of page 886, is NOT correct].

S = So cos2

S = So cos2 = So/2

cos2 = 0.50

cos = 0.707

= 45°

3. (24.3) Locate and describe your own image when you look into a make-up mirror whose radius is 1.2 m when you hold it 25 cm in front of your face.

Locate the image.

Describe the image:

real/virtual; upright/upside-down; smaller/larger

Always start with a clear ray diagram; this is more useful to understanding than just solving the equation. Always ask if the solution to the equation is consistent with your ray diagram.

Now we are ready for the equation,

The focal length is one-half the radius of curvature

f = R/2

f = 120 cm/2 = 60 cm

Your face is the object so the object distance is do = 25 cm

1/di = .01667 - .04

1/di = - 0.02333

di = 1/( - 0.2333)

di = - 43.9 cm

The negative sign means this is a virtual image.

We can find the magnification from

M = - [ ( - 43.0 cm ) / 25 cm ]

M = 1.76

The magnification is M = 1.76. This positive magnification means the image is upright. Since M > 1, the image is enlarged.

4. Consider an object placed 30 cm from a converging lens with a focal length of 10 cm.

a) Draw a ray diagram to locate the image. From this diagram, approximately where is the image located?

b) Characterize the image. Is it real or virtual? Is it upright or upside down? Is it larger or smaller than the object.

c) Now use the image equation to locate the image.

d) Calculate the magnification.

From the ray diagram, we can see that the image is real, inverted, and smaller.

Now, let's do the numerical calculation:

f = 10 cm

do = 30 cm

1/f = 1/do + 1/di

1/di = 1/f - 1/do

1/di = 1/10 cm - 1/30 cm

1/di = 2/30 cm - 1/30 cm

1/di = ( 3 - 1 ) /30 cm

1/di = 2/ 0 cm

1/dii = 1/15 cm

di = 15 cm

Now, the magnification: M = - [ di /do ]

M = - 15 cm / 30 cm

M = - 0.5

The image is inverted and one-half the height of the object.

5. A HeNe laser ( = 634 nm ) shines light onto a double slit with slit separation distance of d = 0.2 mm. An interference pattern of dark and bright areas is produced on a screen 2 m from the double slit. What is the distance between the bright central maximum and the first dark area on either side of it?

A maximum occurs for

d sin = m

But we need the angle for a minimum.

Minima occur for

d sin = (m + 1/2)

This first minimum is for m = 0, so

d sin = (1/2)

(0.2 mm) sin = (0.5)( 634 nm )

( 0.2 x 10 - 3 m) sin = (0.5)( 634 x 10 - 9 m )

sin = ( 0.5)( 634 x 10 - 9 m ) / (0.2 x 10 - 3 m)

sin = ( 0.5)( 634 x 10 - 9 ) / (0.2 x 10 - 3 )

sin = 1.585 x 10 - 3

= 0.0908°

tan = 1.585 x 10 - 3

tan = y/L

y = L tan

y = ( 2.00 m) ( 1.585 x 10 - 3 )

y = 3.17 x 10 - 3 m

y = 3.17 mm

6. As in the diffraction grating lab, a light source consists of a long, thin tube or filament. A meter stick is placed very near the light source. At a distance of L = 2.0 m, there is a diffraction grating with 750 lines / mm.

a) What is the separation distance d between the lines in the grating?

When you look at the light source, where -- how far from the light source, along the meter stick -- will you see the spectrum spread out?

b) At what position will you see red light with  = 700 nm?

c) At what position will you see violet light with  = 400 nm?

First, the diffraction line separation -- the distance between the lines that compose the diffraction grating,

d = 1/N

d = [ 1/750 ] mm

d = 1.33 x 10 - 3 mm

d = 1.33 x 10 - 6 m

Light will be bent by the diffraction grating at angles that correspond to the wavelengths according to

d sin = m

We are concerned only with the first-order spectrum, so m = 1 and

d sin =

d sin =

sin = /d

For red light,

= 700 nm = 700 x 10 - 9 m

sin = (700 x 10 - 9 m)/(1.33 x 10 - 6 m)

sin = (700 x 10 - 9)/(1.33 x 10 - 6)

sin = 0.526

= 31.7°

tan = 0.619

tan = y/L

y = L tan

y = ( 2.00 m) (0.619)

y = 1.23 m = 123 cm

d sin =

sin = /d

For violet light,

= 400 nm = 400 x 10 - 9m

sin = (400 x 10 - 9 m)/(1.33 x 10 - 6 m)

sin = (400 x 10 - 9)/(1.33 x 10 - 6)

sin = 0.301

= 17.5°

tan = 0.315

tan = y/L

y = L tan

y = ( 2.00 m) (0.315)

y = 0.631 m = 63.1 cm

7. A converging lens is placed at the zero mark on an optical bench. An object is placed on the left so that a sharp image is formed on a screen at x = 30 cm. A diverging lens, with focal lenght f = - 30 cm is then placed to the right of the first lens, at x = 10 cm. This causes the image on the screen to be fuzzy. Where must the screen be placed to make the image sharp?

These diagrams may help with question 7.

(seven!?!?! -- I just couldn't pass this one up!)

Well, I did decide to pass this one up. But here's the solution to another interesting problem, nonetheless.

A converging lens is placed at the zero mark on an optical bench.

An object is placed on the left so that a sharp image is formed on a screen at

x = 30 cm.

A diverging lens, with focal lenght f = - 30 cm is then placed to the right of the first lens, at x = 10 cm. This causes the image on the screen to be fuzzy.

Where must the screen be placed to make the image sharp?

The image of the first lens is treated as the object for the second lens.

Since this new "object" is on the right of the lens, the object distance is negative,

do = - 20 cm

1/f = 1/do + 1/di

1/di = 1/f - 1/do

1/di = 1/( - 30 cm) - 1/( - 20 cm)

1/di = - 1/( 30 cm) + 1/( 20 cm)

1/di = - 2/( 60 cm) + 3/( 60 cm)

1/di = (- 2 + 3 )/( 60 cm)

1/di = (+ 1 )/( 60 cm)

1/di = 1/( 60 cm)

di = 60 cm

Since the image distance is positive, we know this is a real image and it can, indeed, be projected onto the screen.

If the image distance is 60 cm, this is 60 cm from the second lens so it will be located at

x = 70 cm

since the second lens is located at x = 10 cm.

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