**PHY1161G
Exam #3
**

**November 1, 2000
**

For reflection,

For refraction, n

n = c/v

image equation

apparent depth d

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 q = m l maxima for m = 0, 1, 2, . . .

diffraction by a **slit**: D = D sin

First **minimum** for D = D sin
=

sin =
/ D

/ D

**Minima** for D = D sin
= **m** , m =
1, 2, 3, . . .

optical resolution:
1.22( / D )

polarization: S = S_{o} cos^{2}

thin films: ’
= / n

2t = m ’ =
m( / n); max, m
= 0, 1, 2, . . . n_{1 }< n_{2 }<
n_{3}

2t = (m+^{1}/_{2}) ’
= (m+^{1}/_{2})(
/ n);

min, m = 0, 1, 2, . . . n_{1 }< n_{2 }<
n_{3}

2t = m ’ =
m( / n); min, m
= 0, 1, 2, . . . n_{1 }< n_{2 }>
n_{3}

2t = (m+^{1}/_{2}) ’
= (m+^{1}/_{2})(l
/ n);

max, m = 0, 1, 2, . . . n_{1 }< n_{2 }>
n_{3
}Simple Magnifier: M_{ang} = ^{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.

Statistics:High: 96

Mean: 65

Low: 18 (then 29)

_{
}

We begin with Snell’s law,

(1.33)(sin 30°) = (1.00) sin

(1.33)(0.5) = (1.00) sin

_{2} =
41.7°

The intensity is given by

[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 = S

cos

cos = 0.707

= 45°

**b) **Consider two ** parallel** polarizing filters,
A and B, as shown below. Another polarizing filter,

The light coming ** from** a polarizer is always aligned
with the direction of the polarizer.

If polarizer **C** is **perpendicular** to polarizer
**A**, no light will get through. If polarizer **C** is
**perpendicular** to polarizer **B**, no light will get
through. But polarizers **A** and **B** are parallel.
Therefore, polarizer **C** will be perpendicular to both **A**
and **B** ** twice** as it is rotated through
360

c)

The light coming ** from** a polarizer is always aligned
with the direction of the polarizer.

If polarizer **C** is **perpendicular** to polarizer
**A**, no light will get through. If polarizer **C** is
**perpendicular** to polarizer **B**, no light will get
through. Polarizers **A** and **B** are perpendicular.
Therefore, polarizer **C** will be perpendicular to either
**A** or **B** ** four times** as it is rotated
through 360

Locate the image.

Describe the image:

Always start with a clear ** ray diagram**; this is more
useful to

** Now** we are ready for the image equation,

The focal length is one-half the radius of curvature

f = 120 cm/2 = 60 cm

Your face is the object so the object distance is d_{o} =
25 cm

1 / d_{i} = .01667 - .04

1 / d_{i} = - 0.02333

d_{i} = 1 / ( - 0.2333)

d_{i} = - 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

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

Locate and characterize the image:

Where is the image located?

Is the image real/virtual; upright/upside-down; smaller/larger?

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

Now, let’s do the numerical calculation:

d

d

Now, the magnification: M = - [
^{d}i / d_{o} ]

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

But we need the angle for a minimum.

Minima occur for

This first minimum is for m = 0, so

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

( 0.2 x 10

sin = ( 0.5)( 634 x 10

sin = ( 0.5)( 634 x 10

sin = 1.585 x 10

= 0.0908

tan = 1.585 x 10

tan =

y = L tan

y = ( 2.00 m) ( 1.585 x 10

y = 3.17 x 10

**6.** As in the lecture demonstration, 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,

d = [

d = 1.33 x 10

d = 1.33 x 10

Unitsareimportant!Don't forget to carefully consider andincludeunits!

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

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

d sin =

sin = ^{
}/_{d}

Forredlight,

sin = (700 x 10

sin = (700 x 10

sin = 0.526

= 31.7

tan = 0.619

tan =

y = ( 2.00 m) (0.619)

y = 1.23 m = 123 cm

d sin =

sin =

For **violet** light,

sin = (400 x 10

sin = (400 x 10

sin = 0.301

= 17.5

tan = 0.315

tan =

y = ( 2.00 m) (0.315)

y = 0.631 m = 63.1 cm