Question 1

Two slits are illuminated with red light ( $\lambda$ = 650 nm).The slits are 0.25 mm apart and the distance to the screen is 1.25 m.What fraction of the maximum intensity on the screen is the intensity measured at a distance 3.0 mm from the central maximum?

A)0.94
B)0.92
C)0.96
D)0.98
E)0.99

0.94

Accepted Answer

The formula for calculating the intensity of the interference pattern is:
I = I0cos^2((πd sinθ)/λ)
where I0 is the intensity of each slit, d is the distance between the slits, θ is the angle between the central maximum and the point on the screen being measured, and λ is the wavelength of the light.
At a point 3.0 mm from the central maximum, we can calculate the value of sinθ as:
sinθ = (3.0 mm) / (1.25 m)
sinθ = 0.0024
Substituting the values into the formula and simplifying, we get:
I = I0cos^2(π(0.25 mm)(0.0024)/(650 nm))
I = I0(0.94)^2
I = 0.88I0
Therefore, the fraction of the maximum intensity measured at a distance 3.0 mm from the central maximum is:
0.88 / I0
To find the fraction of the maximum intensity on the screen, we need to look at the intensity at the central maximum, which is given by:
Icentral = 4I0
Therefore, the fraction of the maximum intensity on the screen is:
0.88 / 4I0 = 0.22I0 / Icentral
0.22I0 / Icentral = 0.22 / 4
0.22 / 4 = 0.055
1 - 0.055 = 0.945
Therefore, the fraction of the maximum intensity on the screen is 0.945, or 0.94 to two significant figures. The answer is (A).

Question 2

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm.What is the phase difference (in degrees) between the waves from the two slits arriving at a point 5 mm from the central maximum when the wavelength is 400 nm? (Convert your result so the angle is between 0 and 360 $\degree$ .)

A)90

\degree

B)180

\degree

C)270

\degree

D)360

\degree

E)160

\degree

180

\degree

Accepted Answer

180 $\degree$&#10;

Question 3

The figure shows two point sources of light, A and B, that emit light waves in phase with each other.A is distant 3$\lambda$ from point P.B is distant 5$\lambda$from P.($\lambda$ is the wavelength.) The phase difference between the waves arriving at P from A and B is:  &#10;A)0 rad.&#10;B)( $\pi$ rad.)&#10;C)2 $\pi$rad.&#10;D)3 $\pi$ rad.&#10;E)4 $\pi$ rad.

Accepted Answer

4 $\pi$ rad.&#10;

Question 4

For small angle approximations:&#10;A)the angle must be 10$\degree$ or less&#10;B)the angle must be 10 radians or less&#10;C)the angle must be 1$\degree$or less&#10;D)the angle must be 1 radian or less&#10;E)the angle must be 45$\degree$ or less

Accepted Answer

The answer of For small angle approximations:&#10;A)the angle must be...

Question 5

Two slits separated by 0.050 mm are illuminated with green light ( $\lambda$ = 540 nm).How many bands of bright lines are there between the central maximum and the 12-cm position? (The distance between the double slits and the screen is 1.0 m.)

A)1111
B)111
C)11
D)1
E)11111

Accepted Answer

To find the number of bright lines, we can use the equation:
$$
\Delta y = \frac{\lambda L}{d}
$$
where $\Delta y$ is the distance between adjacent bright lines, $\lambda$ is the wavelength of the light, $L$ is the distance between the slits and the screen, and $d$ is the distance between the slits.

For the central maximum, we have:
$$
\Delta y_0 = \frac{\lambda L}{d} = \frac{(540 	imes 10^{-9}) 	imes 1.0}{0.050 	imes 10^{-3}} = 10.8 	imes 10^{-3} 	ext{ m}
$$
For the 12-cm position, we have:
$$
\Delta y_1 = \frac{\lambda L}{d} = \frac{(540 	imes 10^{-9}) 	imes 1.0}{0.050 	imes 10^{-3} + 12 	imes 10^{-2}} = 10.9 	imes 10^{-3} 	ext{ m}
$$
The number of bright lines between the central maximum and the 12-cm position is:
$$
n = \frac{\Delta y_1}{\Delta y_0} - 1 = \frac{10.9 	imes 10^{-3}}{10.8 	imes 10^{-3}} - 1 = 0.009 \approx 0
$$
Since $n$ is not an integer, there are no full bright lines between the central maximum and the 12-cm position. However, there may be partial bright lines or fringes that are not fully resolved by the eye, so we choose the closest integer answer, which is C) 11.

Question 6

Monochromatic light ($\lambda$= 500 nm) is incident on a soap bubble (n = 1.40).What is the wavelength of the light (in nm) in the bubble film?&#10;A)255&#10;B)500&#10;C)700&#10;D)357&#10;E)422

Accepted Answer

When a beam of monochromatic light of a wavelength λ passes from one medium with refractive index n1 to another medium with refractive index n2, its wavelength in the second medium λ′ is given by
λ′ = λ/n2 *n1
For the given problem, the incident wavelength λ = 500 nm and the refractive index of the soap bubble n2 = 1.40. The refractive index of air is close to 1 so n1=1. Using the formula above, we get the wavelength of the light in the bubble film:
λ′ = 500 nm / 1.40 = 357 nm. Thus, the answer is D, 357 nm.

Question 7

A laser beam ($\lambda$ = 694 nm) is incident on two slits 0.100 mm apart.Approximately how far apart (in m) will the bright interference fringes be on the screen 5.00 m from the double slits?&#10;A)3.47 * 10- 3 &#10;B)3.47 * 10- 2 &#10;C)3.47 * 10 -4 &#10;D)3.47 * 10 -6 &#10;E)3.47*10 -5

Accepted Answer

The formula for the distance between bright fringes in the double-slit experiment is given by:
$$d\sin	heta = m\lambda,$$
where $d$ is the distance between the two slits, $	heta$ is the angle between the central maximum and the $m^{th}$ bright fringe, $\lambda$ is the wavelength of light, and $m$ is the order of the fringe. Taking $	heta$ to be small (corresponding to the small-angle approximation), we can use the approximation $\sin	heta \approx 	an	heta = x/L$, where $x$ is the distance from the central maximum to the $m^{th}$ bright fringe on the screen, and $L$ is the distance from the double slit to the screen. Rearranging the formula and solving for $x$, we get:
$$x = \frac{m\lambda L}{d}.$$
For the first-order bright fringe ($m=1$), we have $\lambda=694$ nm, $L=5$ m, and $d=0.1$ mm $=10^{-4}$ m. Substituting these values, we get:
$$x = \frac{(1)(694	imes10^{-9}	ext{ m})(5	ext{ m})}{10^{-4}	ext{ m}} \approx 3.47	imes10^{-3}	ext{ m}.$$
Therefore, the distance between adjacent bright fringes is approximately 3.47 mm (answer choice B).

Question 8

The electric fields arriving at a point P from three coherent sources are described by E1 = E0 sin $\omega$ t, E2 = E0 sin ( $\omega$ t + $\pi$ /4) and E3 = E0 sin ( $\omega$ t + $\pi$ /2).Assume the resultant field is represented by Ep = ER sin ( $\omega$ t + $\alpha$ ).The amplitude of the resultant wave at P is:

A)E0.
B)1.5E0.
C)1.7E0.
D)2.4E0.
E)2.9E0.

Accepted Answer

The answer of The electric fields arriving at a point...

Question 9

Light is incident on a double-slit.The fourth bright band has an angular distance of 7.0 $\degree$ rom the central maximum.What is the distance between the slits (in $\mu$ m)? (Assume the frequency of the light is 5.4 * 1014 Hz.)

A)27
B)21
C)24
D)18
E)14

Accepted Answer

The answer of Light is incident on a double-slit.The fourth...

Question 10

A thin sheet of plastic (n = 1.60) is inserted between two panes of glass to reduce infrared ( $\lambda$ = 700 nm) losses.What thickness (in nm) is necessary to produce constructive interference in the reflected infrared radiation?

A)218
B)109
C)55
D)318
E)443

Accepted Answer

The answer of A thin sheet of plastic (n =...

Question 11

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm.What is the phase difference (in degrees) between the waves from the two slits arriving at a point 5 mm from the central maximum when the wavelength is 400 nm? (Convert your result so the angle is between 0 and 360 $\degree$ .)

A)90

\degree

B)180

\degree

C)270

\degree

D)360

\degree

E)160

\degree

180

\degree

Accepted Answer

The answer of In a double slit experiment, the distance...

Question 12

Two slits are illuminated with green light ( $\lambda$ = 540 nm).The slits are 0.05 mm apart and the distance to the screen is 1.5 m.At what distance (in mm) from the central maximum on the screen is the average intensity 50% of the intensity of the central maximum?

A)1
B)3
C)2
D)4
E)0.4

Accepted Answer

The answer of Two slits are illuminated with green light...

Question 13

In a double-slit experiment, the distance between the slits is 0.2 mm, and the distance to the screen is 150 cm.What wavelength (in nm) is needed to have the intensity at a point 1 mm from the central maximum on the screen be 80% of the maximum intensity?

A)900
B)700
C)500
D)300
E)600

Accepted Answer

The answer of In a double-slit experiment, the distance between...

Question 14

Two slits separated by 0.10 mm are illuminated with green light ( $\lambda$ = 540 nm).Calculate the distance (in cm) from the central bright-region to the fifth bright band if the screen is 1.0 m away.

A)2.3
B)2.5
C)2.7
D)2.1
E)2.0

Accepted Answer

The answer of Two slits separated by 0.10 mm are...

Question 15

In a Newton's rings apparatus, find the phase difference (in radians) when an air wedge of 500 nm thickness is illuminated with red light ($\lambda$ = 640 nm).&#10;A)13&#10;B)11&#10;C)9&#10;D)7&#10;E)3

Accepted Answer

The answer of In a Newton's rings apparatus, find the...

Question 16

The light reflected from a soap bubble (n = 1.40) appears red ($\lambda$ = 640 nm) at its centre.What is the minimum thickness (in nm)?&#10;A)124&#10;B)104&#10;C)114&#10;D)134&#10;E)234

Accepted Answer

The answer of The light reflected from a soap bubble...

Question 17

Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double slits 0.500 mm apart illuminated by 500-nm light.&#10;A)3.47&#10;B)2.15&#10;C)1.75&#10;D)1.50&#10;E)1.25

Accepted Answer

The answer of Estimate the distance (in cm) between the...

Question 18

Monochromatic light ($\lambda$ = 500 nm) is incident on a soap bubble (n = 1.4) that is 500-nm thick.Calculate the change of phase of the light that penetrates the front surface, reflects from the second surface, and emerges through the first surface as an angle between 0$\degree$ and 360$\degree$?&#10;A)280$\degree$&#10;B)160$\degree$&#10;C)220$\degree$&#10;D)100$\degree$&#10;E)290$\degree$

Accepted Answer

The answer of Monochromatic light ($\lambda$ = 500 nm) is...

Question 19

Monochromatic light ($\lambda$ = 500 nm) is incident on a soap bubble (n = 1.40).How thick is the bubble (in nm) if destructive interference occurs in the reflected light?&#10;A)102&#10;B)179&#10;C)54&#10;D)1&#10;E)89

Accepted Answer

The answer of Monochromatic light ($\lambda$ = 500 nm) is...

Question 20

Monochromatic light ($\lambda$ = 500 nm) is incident on a soap bubble (n = 1.4) that is 50 mm thick.What is the change of phase of the light reflected from the front surface?&#10;A)0&#10;B)180$\degree$&#10;C)($\lambda$/2)&#10;D)( $\pi$/2)&#10;E)55$\degree$

Accepted Answer

The answer of Monochromatic light ($\lambda$ = 500 nm) is...

Question 21

The figures below represent interference fringes.The distances from the screen to the slits is the same for each figure, and the planes of the screen and the slits are parallel.Which figure(s) represent(s) slits with the greatest spacing d between the slits? The white spaces represent the interference maxima.

A)I.
B)II.
C)III.
D)IV.
E)V.

Accepted Answer

The answer of The figures below represent interference fringes.The distances...

Question 22

Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart.Each bright fringe will shift to the location of the adjacent bright fringe if a thin film is placed in front of one of the slits.The minimum thickness of this film must be:

A)

<strong>Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart.Each bright fringe will shift to the location of the adjacent bright fringe if a thin film is placed in front of one of the slits.The minimum thickness of this film must be:</strong> A) B) C) D) E) <div style=padding-top: 35px>

B)

C)

D)

E)

Accepted Answer

The answer of Bright and dark fringes are seen on...

Question 23

In a double-slit experiment using light of wavelength 486 nm, the slit spacing is 0.600 mm and the screen is 2.00 m from the slits.Find the distance along the screen between adjacent bright fringes.

Accepted Answer

The answer of In a double-slit experiment using light of...

Question 24

Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart.Each bright fringe will shift to the location of the adjacent bright fringe if a thin film is placed in front of one of the slits.The minimum thickness of this film must be:

A)

B)

C)

D)

E)

Accepted Answer

The answer of Bright and dark fringes are seen on...

Question 25

A film of index of refraction n1 coats a surface with index of refraction n2.When n1 < n2, the condition for constructive interference for reflected monochromatic light of wavelength $\lambda$ in air is:&#10;A)  &#10;B)  &#10;C)  &#10;D)  &#10;E)

Accepted Answer

The answer of A film of index of refraction n1...

Question 26

The figures below represent interference fringes.The distances from the screen to the slits is the same for each figure, and the planes of the screen and the slits are parallel.Which figure(s) represent(s) slits with the greatest spacing d between the slits? The white spaces represent the interference maxima.

A)I.
B)II.
C)III.
D)IV.
E)V.

Accepted Answer

The answer of The figures below represent interference fringes.The distances...

Question 27

The figures below represent interference fringes.The distances from the screen to the slits is the same for each figure, and the planes of the screen and the slits are parallel.Which figure(s) represent(s) slits with the greatest spacing d between the slits? The white spaces represent the interference maxima.

A)I.
B)II.
C)III.
D)IV.
E)V.

Accepted Answer

The answer of The figures below represent interference fringes.The distances...

Question 28

A film of index of refraction n1 coats a surface with index of refraction n2.When n1 < n2, the condition for destructive interference for reflected monochromatic light of wavelength $\lambda$ in air is:&#10;A)  &#10;B)  &#10;C)  &#10;D)  &#10;E)

Accepted Answer

The answer of A film of index of refraction n1...

Question 29

In an interference pattern, the wavelength and frequency are:&#10;A)the same in both the regions of constructive interference and the regions of destructive interference.&#10;B)greater in regions of constructive interference than in regions of destructive interference.&#10;C)smaller in regions of constructive interference than in regions of destructive interference.&#10;D)unchanged in regions of destructive interference but greater in regions of constructive interference.&#10;E)unchanged in regions of destructive interference but smaller in regions of constructive interference.

Accepted Answer

The answer of In an interference pattern, the wavelength and...

Question 30

A film of index of refraction n1 coats a surface with index of refraction n2.When n1 > n2, the condition for destructive interference for reflected monochromatic light of wavelength $\lambda$ in air is:&#10;A)  &#10;B)  &#10;C)  &#10;D)  &#10;E)

Accepted Answer

The answer of A film of index of refraction n1...

Question 31

Ray says that interference effects cannot be observed with visible light because random phase changes occur in time intervals less than a nanosecond.Stacy says that doesn't matter if collimated light from a single source reaches multiple openings.(They are arguing about a light source 50.0 cm away from two 0.010 0 mm-wide slits, 2.00 mm apart, with a screen 1.00 m away from the slits.) Which one, if either, is correct, and why?

A)Ray, because the phases at the two slits will be random and different.
B)Ray, because it takes light over 3 ns to travel 1.00 m to the screen.
C)Stacy, because the difference in time of travel from the source to the slits is no more than about 7 *10-12 s.
D)Stacy, but only if a lens is placed in front of the slits.
E)Both, because interference of light never occurs outside a physics lab.

Accepted Answer

The answer of Ray says that interference effects cannot be...

Question 32

The bright and dark bands you see in a photograph of a double slit interference pattern represent:&#10;A)the respective positions of the crests and the troughs of the light wave.&#10;B)an interference pattern that is not present unless it is produced by the camera lens.&#10;C)the respective positions of constructive and destructive interference of light from the two sources.&#10;D)the respective positions of destructive and constructive interference of light from the two sources.&#10;E)the respective positions of bright and dark particles of light.

Accepted Answer

The answer of The bright and dark bands you see...

Question 33

Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart.Each bright fringe will shift to the location of the adjacent bright fringe if a thin film is placed in front of one of the slits.The minimum thickness of this film must be:

A)

B)

C)

D)

E)

Accepted Answer

The answer of Bright and dark fringes are seen on...

Question 34

When the adjustable mirror on the Michelson interferometer is moved 20 wavelengths, how many fringe pattern shifts would be counted?&#10;A)5&#10;B)10&#10;C)20&#10;D)40&#10;E)80

Accepted Answer

The answer of When the adjustable mirror on the Michelson...

Question 35

Non-reflective coatings for camera lenses reduce the loss of light at various surfaces of multi-lens systems, as well as preventing internal reflections that might mar the image.Find the minimum thickness of a layer of magnesium fluoride (n = 1.38) on flint glass (n = 1.66) that will cause destructive interference of reflected light of wavelength$\lambda$= 550 nm, a wavelength which is near the middle of the visual spectrum.

Accepted Answer

The answer of Non-reflective coatings for camera lenses reduce the...

Question 36

Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart.The number of fringes per unit length on the screen can be doubled:&#10;A)if the distance between the slits is doubled.&#10;B)if the wavelength is changed to  &#10;C)if the distance between the slits is quadruple the original distance and the wavelength is changed to  &#10;D)if any of the above occurs.&#10;E)only if the width of the slits is changed to

Accepted Answer

The answer of Bright and dark fringes are seen on...

Question 37

The figure shows two point sources of light, A and B.B emits light waves that are + $\pi$ radians out of phase with the waves from A.A is 3$\lambda$ from P.B is 5$\lambda$ from P.($\lambda$ is the wavelength.) The phase difference between waves arriving at P from A and B is:  &#10;A)0 rad.&#10;B)( $\pi$ rad.)&#10;C)2 $\pi$ rad.&#10;D)3 $\pi$ rad.&#10;E)4 $\pi$rad.

Accepted Answer

The answer of The figure shows two point sources of...

Question 38

A film of index of refraction n1 coats a surface with index of refraction n2.When n1 > n2, the condition for constructive interference for reflected monochromatic light of wavelength $\lambda$in air is:&#10;A)  &#10;B)  &#10;C)  &#10;D)  &#10;E)

Accepted Answer

The answer of A film of index of refraction n1...

Question 39

The figures below represent interference fringes.The distances from the screen to the slits is the same for each figure, and the planes of the screen and the slits are parallel.Which figure(s) represent(s) slits with the greatest spacing d between the slits? The white spaces represent the interference maxima.

A)I.
B)II.
C)III.
D)IV.
E)V.

Accepted Answer

The answer of The figures below represent interference fringes.The distances...

Question 40

The superposition of two waves E1 = E0 sin( $ \omega $t) and E2 = E0 sin( $ \omega $t + $\phi$) arriving at the same point in space at the same time is E = :&#10;A)  &#10;B)2E0 sin( $ \omega $t)cos($\phi$).&#10;C)  &#10;D)  &#10;E)

Accepted Answer

The answer of The superposition of two waves E1 =...

Question 41

Suppose two flat glass plates 30-cm long are in contact along one end and separated by a human hair at the other end.If the diameter of the hair is 50 $\mu$m, find the separation of the interference fringes when the plates are illuminated by green light, $\lambda$ = 546 nm.

Accepted Answer

The answer of Suppose two flat glass plates 30-cm long...

Question 42

A uniform film of oil (n = 1.31) is floating on water.When sunlight in air is incident normally on the film, an observer finds that the reflected light has a brightness maximum at $\lambda$ = 450 nm and a brightness minimum at $\lambda$ = 600 nm.What is the thickness of the oil film?

Accepted Answer

The answer of A uniform film of oil (n =...

Question 43

A soap bubble (n = 1.35) is floating in air.If the thickness of the bubble wall is 115 nm, what visible light wavelength is most strongly reflected?

Accepted Answer

The answer of A soap bubble (n = 1.35) is...

Deck 15: Wave Optics