Question 1

You illuminate a slit 0.20 mm wide with monochromatic light of wavelength 590 nm, and focus the Fraunhofer diffraction pattern on a screen with a lens of focal length
2)00 m. The distance between the two dark fringes on each side of the central bright fringe is
A) 0.295 cm
B) 0.590 cm
C) 1.18 cm
D) 2.95 cm
E) 5.90 cm

Accepted Answer

The distance between the two dark fringes on each side of the central bright fringe can be calculated using the formula:
D = (λL)/d
Where D is the distance between the two dark fringes, λ is the wavelength of the light, L is the distance between the slit and the screen (in this case, the focal length of the lens), and d is the width of the slit.
Plugging in the given values, we get:
D = (590 nm)*(200 m)/(0.20 mm) = 1.18 cm
Therefore, the best choice is C.

Question 2

The white circles in the figure represent minima of the diffraction pattern formed when a bright point-source object is viewed through a small circular opening. In accordance with Rayleigh's criterion, the closest central maximum of another point source that could lie in this pattern and still be barely resolvable
A) would be at point A.
B) would be at point B.
C) would be at point C.
D) would be at point D.
E) must lie outside all discernible diffraction rings.

Accepted Answer

The answer of  The white circles in the figure...

Question 3

A diffraction grating has 5000 lines per centimeter and is 1.5 centimeters wide. When radiation is incident upon the grating a second-order maximum is observed at an angle of 37º. The wavelength of this light is
A) 400 nm
B) 600 nm
C) 900 nm
D) 1200 nm
E) 2400 nm

Accepted Answer

The wavelength ($\lambda$) can be found using the grating equation $d\sin(\theta) = m\lambda$, where $d$ is the grating spacing, $\theta$ is the diffraction angle, and $m$ is the order of the maximum. Given 5000 lines per cm, the spacing $d = \frac{1}{5000 \text{ lines/cm}} = \frac{1}{500000 \text{ lines/m}} = 2 \times 10^{-6} \text{ m}$. For the second-order maximum ($m = 2$) at $37^\circ$, solving for $\lambda$ gives $\lambda = \frac{d\sin(\theta)}{m} = \frac{2 \times 10^{-6} \text{ m} \times \sin(37^\circ)}{2} \approx 600 \text{ nm}$.

Question 4

Green light of wavelength 500 nm is incident normally on a diffraction grating that has 5300 lines/cm. The second-order image is diffracted at an angle from the normal of
A) 10º
B) 16º
C) 32º
D) 48º
E) 57º

Accepted Answer

The equation for diffraction on a grating is: 
d sinθ = mλ
where 
d = distance between adjacent slits
θ = angle of diffraction
m = order of diffraction
λ = wavelength of light

Using the given values:
d = 1 / (5300 lines/cm) = 1.8868 x 10^-4 cm
m = 2
λ = 500 nm = 5 x 10^-5 cm

Substituting into the equation, we get:
d sinθ = mλ
1.8868 x 10^-4 cm x sinθ = 2 x 5 x 10^-5 cm
sinθ = 0.53
θ = sin^-1(0.53) = 32º

Therefore, the answer is C) 32º.

Question 5

An antenna for receiving television signals from a satellite has a diameter of 3 m. The signals have a wavelength of 24 cm. If the diameter of the antenna is reduced to 50 cm, for what signal wavelengths does the antenna still have the same angular resolution?
A) 4.0 cm
B) 6.0 cm
C) 8.0 cm
D) 1.4 m
E) 24 cm

Accepted Answer

For an antenna to have the same angular resolution, the ratio of its diameter (D) to the wavelength (λ) should be constant. This is commonly referred to as the Rayleigh criterion. 
So, we can write:
D1/λ1 = D2/λ2
Where D1 = 3 m, λ1 = 24 cm = 0.24 m, and D2 = 0.5 m (reduced diameter).
We can solve for the wavelength that satisfies the above equation:
λ2 = (D2/D1)λ1 = (0.5/3) x 0.24 = 0.04 m = 4.0 cm
Therefore, the antenna will have the same angular resolution for signals with a wavelength of 4.0 cm. 
The other wavelengths given in the answer choices are either too small or too large, and therefore, the antenna will not have the same angular resolution for those wavelengths.

Question 6

For a grating with d = 3.5$\lambda$, the maximum order m of an interference maximum that can be observed for a specified $\lambda$ is
A) 3.5
B) 3
C) 1
D) 2
E) 2.5

Accepted Answer

The answer of For a grating with d = 3.5$\lambda$,...

Question 7

For a given grating, it is observed that a third-order line (m = 3) with $\lambda$ = 465.3 nm overlaps a second-order line (has the same $\theta$ as a line with m = 2). The wavelength of the second-order line is
A) 304 nm
B) 698 nm
C) 913 nm
D) 152 nm
E) 372 nm

Accepted Answer

Let's use the grating equation:

mλ = d(sinθ + sinϕ)

Where m is the order, λ is the wavelength, d is the grating spacing, θ is the angle of incidence, and ϕ is the angle of diffraction.

For the third-order line:

3(465.3 nm) = d(sinθ + sinϕ1)

For the second-order line:

2λ = d(sinθ + sinϕ2)

We know that both lines have the same F1F1F1S1θF1F1F10 (angle of incidence), so we can equate the values:

sinθ + sinϕ1 = sinθ + sinϕ2

sinϕ1 = sinϕ2

Now we can substitute sinϕ1 for sinϕ2 in the second equation:

2λ = d(sinθ + sinϕ1)

2λ = 3(465.3 nm)

λ = 697.95 nm

The closest answer choice is B, 698 nm.

Question 8

You illuminate a slit 0.20 mm wide with monochromatic light of wavelength 600 nm, and focus the Fraunhofer diffraction pattern on a screen with a lens of focal length
1)50 m. The distance on the screen from the center of the central maximum to the first minimum is
A) 0.675 cm
B) 0.600 cm
C) 0.300 cm
D) 0.450 cm
E) 0.225 cm

Accepted Answer

The answer of You illuminate a slit 0.20 mm wide...

Question 9

The type of electromagnetic radiation that is employed when crystals are used as diffraction gratings is
A) infrared
B) Xrays
C) visible
D) ultraviolet
E) microwaves

Accepted Answer

Xrays have a wavelength small enough to interact with the interatomic spacings in crystals, providing information on their structure through diffraction.

Question 10

The size of the smallest things that can be seen with an optical microscope is limited by diffraction. Which of the following could help a microscopist see smaller things?
A) A more powerful microscope could be used.
B) The microscope could have a lens with a shorter focal length.
C) The microscope could have a lens with a longer focal length.
D) The diameter of the lens could be smaller.
E) Light with a shorter wavelength could be used.

Accepted Answer

The size limit of the smallest things that can be seen with an optical microscope is proportional to the wavelength of the light used. Therefore, using light with a shorter wavelength would enable a microscopist to see smaller things. Increasing the power or adjusting the lens would not change the fundamental limit imposed by diffraction.

Question 11

The 200-in reflecting telescope at Mt. Palomar has an objective mirror with a focal length of 16.8 m. The effective aperture is the diameter of the 200-in mirror (200 in = 5.08 m). For light in the blue region (450 nm), what is the smallest angle between two stars that this telescope can ideally resolve? (In practice, the limit of resolution is due to atmospheric turbulence.) A) 1.1* 10^-8 rad B) 1.1 * 10^-7 rad C) 7.3 * 10^-7 rad D) 1.1 *10^-4 rad E) 8.9 * 10^-5 rad

Accepted Answer

The smallest angle that can be resolved by a telescope is given by the Rayleigh criterion:

θ = 1.22 λ/D

where θ is the smallest angle that can be resolved, λ is the wavelength of light, and D is the diameter of the objective mirror (i.e. the effective aperture).

Substituting the values given in the problem statement, we get:

θ = 1.22 (450 nm)/(5.08 m) = 1.1 x 10^-7 radians

Therefore, the answer is B, which is the closest to this value in scientific notation.

Question 12

In accordance with the Rayleigh criterion, two points can be just resolved if the centers of their diffraction patterns are separated by
A) one wavelength.
B) twice the width of either central maximum.
C) one-half the width of either central maximum.
D) the width of the aperture.
E) the reciprocal of one wavelength.

Accepted Answer

According to Rayleigh's criterion, two points can be resolved if the central maximum of one point coincides with the first minimum of the other point. This occurs when the centers of their diffraction patterns are separated by one-half the width of either central maximum.

Question 13

A monochromatic beam of light of wavelength 600 nm falls on a grating at normal incidence and produces a second-order image at an angle of 30º. The grating spacing must be
A) 0.60 µm
B) 2.4 µm
C) 0.30 µm
D) 1.4 µm
E) 1.0 µm

Accepted Answer

The answer of A monochromatic beam of light of wavelength...

Question 14

Diffraction occurs when light passes
A) by a small particle.
B) through a small hole.
C) through a double slit.
D) by a sharp edge.
E) Diffraction occurs in all of these conditions.

Accepted Answer

Diffraction occurs when light passes by a small particle, through a small hole, through a double slit, or by a sharp edge. It is a wave phenomenon that causes interference patterns and affects the directionality and spreading of light.

Question 15

The headlights of an oncoming car are 1.2 m apart. What is the maximum distance from the car at which you can resolve the lights as two sources if the diameter of the pupil of your eye is 5.0 mm and the wavelength of the light is 555 nm?
A) 8.9 km
B) 22 km
C) 4.4 km
D) 5.4 km
E) 13 km

Accepted Answer

The resolution of two sources as distinct depends on the Rayleigh criterion, which is given by the formula $\theta = 1.22 \frac{\lambda}{D}$, where $\theta$ is the angular resolution in radians, $\lambda$ is the wavelength of light, and $D$ is the diameter of the aperture (in this case, the pupil of the eye). To find the maximum distance ($L$) at which the car's headlights can be resolved, we use the formula $\theta = \frac{s}{L}$, where $s$ is the separation between the headlights. Combining the two gives $L = \frac{sD}{1.22\lambda}$.Given: $s = 1.2$ m, $D = 5.0$ mm = $0.005$ m, and $\lambda = 555$ nm = $555 \times 10^{-9}$ m.Substituting these values gives:$$L = \frac{1.2 \times 0.005}{1.22 \times 555 \times 10^{-9}} \approx 8900 \, \text{m} = 8.9 \, \text{km}.$$

Question 16

For a grating with d = 4.5$\lambda$, the maximum order m of an interference maximum that can be observed for a specified $\lambda$ is
A) 4.5
B) 3
C) 1
D) 2
E) 4

Accepted Answer

The answer of For a grating with d = 4.5$\lambda$,...

Question 17

Rayleigh's criterion is most closely associated with
A) diffraction
B) coherence
C) dispersion
D) polarization
E) reflection

Accepted Answer

Rayleigh's criterion deals with the resolution of two closely spaced objects and is used to determine the smallest separation between two points that produce distinct and separate images. It is most commonly used in the context of diffraction-limited optical systems.

Question 18

The diffraction image of the point object O is indicated in the figure. Of the point objects 1, 2, 3, 4, and 5, the one closest to O whose diffraction image could just be resolved from that of O according to the Rayleigh criterion is
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The answer of  The diffraction image of the point...

Question 19

The pupil of the human eye has a diameter of about 5 mm. When the wavelength of light incident on the pupil is 500 nm, the smallest angular separation of two resolvable sources is approximately
A) 1"
B) 1'
C) 1º
D) 10º
E) 1 radian

Accepted Answer

The smallest angular separation of two resolvable sources is given by the Rayleigh criterion: 
θ = 1.22 λ/D 
where λ is the wavelength of light, D is the diameter of the pupil, and θ is the angular separation of the sources. 
Plugging in the values, we get: 
θ = 1.22 (500 x 10^-9 m) / 5 x 10^-3 m 
θ = 1.22 x 10^-4 radians 
Converting to arcminutes (1 radian = 60 arcminutes), we get: 
θ = 7.32 arcminutes 
So the smallest angular separation of two resolvable sources is approximately 7 arcminutes, which is closest to answer B: 1'.

Question 20

In demonstrating single-slit Fraunhofer diffraction, decreasing the wavelength of the light while keeping the slit width constant
A) does not affect the width of the central maximum.
B) increases the number of fringes within the central maximum.
C) decreases the number of fringes within the central maximum.
D) increases the width of the central maximum.
E) decreases the width of the central maximum

Accepted Answer

The width of the central maximum is given by the equation w = 2λL/w, where λ is the wavelength of the light, L is the distance from the slit to the screen, and w is the width of the slit. As the wavelength of the light decreases, the width of the central maximum decreases. This is because the smaller wavelength causes the diffraction pattern to spread out less, resulting in a narrower central maximum. Therefore, decreasing the wavelength of the light while keeping the slit width constant will decrease the width of the central maximum.

Quiz 13: Interference and Diffraction

You illuminate a slit 0.20 mm wide with monochromatic light of wavelength 590 nm, and focus the Fraunhofer diffraction pattern on a screen with a lens of focal length 2) 00 m. The distance between the two dark fringes on each side of the central bright fringe is

A diffraction grating has 5000 lines per centimeter and is 1.5 centimeters wide. When radiation is incident upon the grating a second-order maximum is observed at an angle of 37º. The wavelength of this light is

Green light of wavelength 500 nm is incident normally on a diffraction grating that has 5300 lines/cm. The second-order image is diffracted at an angle from the normal of

An antenna for receiving television signals from a satellite has a diameter of 3 m. The signals have a wavelength of 24 cm. If the diameter of the antenna is reduced to 50 cm, for what signal wavelengths does the antenna still have the same angular resolution?

For a grating with d = 3.5 $\lambda$ , the maximum order m of an interference maximum that can be observed for a specified $\lambda$ is

For a given grating, it is observed that a third-order line (m = 3) with $\lambda$ = 465.3 nm overlaps a second-order line (has the same $\theta$ as a line with m = 2) . The wavelength of the second-order line is

You illuminate a slit 0.20 mm wide with monochromatic light of wavelength 600 nm, and focus the Fraunhofer diffraction pattern on a screen with a lens of focal length 1) 50 m. The distance on the screen from the center of the central maximum to the first minimum is

The type of electromagnetic radiation that is employed when crystals are used as diffraction gratings is

The size of the smallest things that can be seen with an optical microscope is limited by diffraction. Which of the following could help a microscopist see smaller things?

In accordance with the Rayleigh criterion, two points can be just resolved if the centers of their diffraction patterns are separated by

A monochromatic beam of light of wavelength 600 nm falls on a grating at normal incidence and produces a second-order image at an angle of 30º. The grating spacing must be

Diffraction occurs when light passes

The headlights of an oncoming car are 1.2 m apart. What is the maximum distance from the car at which you can resolve the lights as two sources if the diameter of the pupil of your eye is 5.0 mm and the wavelength of the light is 555 nm?

For a grating with d = 4.5 $\lambda$ , the maximum order m of an interference maximum that can be observed for a specified $\lambda$ is

Rayleigh's criterion is most closely associated with

The pupil of the human eye has a diameter of about 5 mm. When the wavelength of light incident on the pupil is 500 nm, the smallest angular separation of two resolvable sources is approximately

In demonstrating single-slit Fraunhofer diffraction, decreasing the wavelength of the light while keeping the slit width constant