Question 1

Two point sources of light of wavelength 600 nm are being viewed by a telescope of aperture width 2.0 cm.What is the closest angular separation at which the two sources can be resolved? A) 7.3 $\times$ 10^-5 rad B) 3.7 $\times$ 10^-5 rad C) 7.3 $\times$ 10^-5 deg D) 3.7 $\times$ 10^-5 deg E) 4.2 $\times$ 10^-3 deg

Accepted Answer

The angular resolution of a telescope can be approximated by the Rayleigh criterion, which is given by:
θ = 1.22λ/D
where θ is the minimum angular separation that can be resolved, λ is the wavelength of light, and D is the diameter of the aperture of the telescope.
Substituting the given values, we get:
θ = 1.22(600 nm)/(2.0 cm)
θ = 3.7 x 10^-5 radians
Converting to degrees, we get:
θ = 2.1 x 10^-3 degrees
Therefore, the closest angular separation at which the two sources can be resolved is approximately 3.7 x 10^-5 radians or 2.1 x 10^-3 degrees, which is closest to option B.

Question 2

Holograms appear three dimensional because
A) they are in color.
B) they are transparent.
C) they consist of Young double slits.
D) they recreate wave fronts as if they came from the original source.
E) of their spectral response to refraction.

Accepted Answer

Holograms recreate wavefronts and diffracted light patterns as if they originated from the original source. This gives the appearance of a three-dimensional image.

Question 3

If the eye can resolve points separated by 2.20 $\times$ 10^-4 rad,when viewed at 12.0 inches,how many dpi is this? (dpi = dots per inch) A) 250 B) 340 C) 380 D) 400 E) 590

Accepted Answer

The answer of If the eye can resolve points separated...

Question 4

A single slit of width 0.030 mm is used to project a diffraction pattern of 500 nm light on a screen at a distance of 2.00 m from the slit.What is the width of the central maximum?
A) 1.7 cm
B) 3.3 cm
C) 6.7 cm
D) 1.5 mm
E) 3.0 mm

Accepted Answer

The answer of A single slit of width 0.030 mm...

Question 5

The resolution of a microscope can be most improved by
A) increasing the wavelength and diameter of the lens.
B) decreasing the wavelength and the diameter of the lens.
C) decreasing the wavelength and increasing the diameter of the lens.
D) increasing the wavelength and decreasing the diameter of the lens.

Accepted Answer

The resolution of a microscope can be improved by decreasing the wavelength of the imaging light and increasing the diameter of the lens. This is because a smaller wavelength of light allows for greater magnification and a larger lens diameter allows for more light gathering and hence greater resolution. Therefore, option C is the best choice.

Question 6

What is the angular resolution of the eye at a wavelength of 500 nm for a pupil diameter of 2.8 mm? A) 1.8 $\times$ 10^-4 rad B) 2.2 $\times$ 10^-4 rad C) 8.9 $\times$ 10^-5 rad D) 4.5 $\times$ 10^-5 rad E) 3.3 $\times$ 10^-5 rad

Accepted Answer

The answer of What is the angular resolution of the...

Question 7

X-rays of fixed wavelength yield a first-order maximum at a grazing angle of 12$^{\circ}$.At what angle would one expect to find the corresponding 3^rd order maximum? A) 4.0$^{\circ}$ B) 12$^{\circ}$ C) 36$^{\circ}$ D) 39$^{\circ}$ E) more information is needed

Accepted Answer

The angle for the nth order maximum is given by the equation nλ = 2d sin(θ). For the 3rd order maximum, the angle is 3 times the angle for the 1st order maximum, so 3 * 12° = 36°. However, due to the periodic nature of the sine function, the angle is slightly larger, making it 39°.

Question 8

White light is shone on a very thin layer of mica (n = 1.57).When the mica layer is observed from the illuminated side,interference maxima for wavelengths of blue (480 nm) and yellow (560 nm)are observed.What is the thickness of the mica layer?
A) 1.58 $\mu$m
B) 0.79 $\mu$m
C) 3.16 $\mu$m
D) insufficient information is given.

Accepted Answer

The thickness of the mica layer is equal to one-half of the wavelength of the light in the medium. For blue light, the wavelength in the medium is 480 nm / 1.57 = 306 nm. For yellow light, the wavelength in the medium is 560 nm / 1.57 = 357 nm. The thickness of the mica layer is therefore (306 nm + 357 nm) / 2 = 1.58 µm.

Question 9

Max von Laue is known for his work with
A) double-slit interference.
B) single-slit diffraction.
C) x-ray crystallography.
D) holography.
E) magnetic resonance imaging.

Accepted Answer

Max von Laue is primarily known for his work in x-ray crystallography which won him the Nobel Prize in Physics in 1914. He discovered that crystals can diffract X-rays, leading to the development of the technique of X-ray crystallography, which is used to determine the three-dimensional structure of molecules.

Question 10

For a plane wave approaching an obstacle with a rectangular opening,diffraction becomes important in which of the following circumstances?
A) the beam is coherent
B) one or more dimensions of the rectangular opening is near the wavelength of the source
C) both dimensions of the rectangular opening are larger than the wavelength of the source
D) only if the opening is circular

Accepted Answer

Diffraction becomes significant when the dimensions of the opening are comparable to or smaller than the wavelength of the source. Therefore, if one or more dimensions of the rectangular opening is near the wavelength of the source, diffraction will occur. Coherence of the beam or the shape of the opening (circular or rectangular) is not relevant to this phenomenon.

Question 11

Holography was invented by
A) Max von Laue.
B) Sir David Brewster.
C) Rosalind Franklin.
D) Dennis Gabor.
E) Bill Gates.

Accepted Answer

Holography was invented by Dennis Gabor in 1947, who received the Nobel Prize in Physics in 1971 for his work on the principles of holography. The other options listed are not associated with the invention of holography.

Question 12

In a single-slit experiment,light of wavelength 500 nm is used to produce a diffraction pattern on a screen 4.0 m away.If the distance between both 3^rd order minima is 6.0 cm,what is the slit width? A) 2.0 $\times$ 10^-4 m B) 1.0 $\times$ 10^-4 m C) 4.0 $\times$ 10^-4 m D) 0.67 $\times$ 10^-4 m E) 0.44 $\times$ 10^-4 m

Accepted Answer

The distance between the 3rd order minima is given by:$$d\sin	heta=3\lambda$$where d is the slit width, 	heta is the angle of diffraction, and \lambda is the wavelength of light.We can use the small angle approximation to get:$$	an	heta=\frac{y}{L}$$where y is the distance from the central maximum to the 3rd order minima and L is the distance from the slit to the screen.Substituting this into the first equation, we get:$$d\frac{y}{L}=3\lambda$$Solving for d, we get:$$d=\frac{3\lambda L}{y}$$Plugging in the given values, we get:$$d=\frac{3(500	imes10^{-9}	ext{ m})(4.0	ext{ m})}{0.060	ext{ m}}=1.0	imes10^{-4}	ext{ m}$$

Question 13

If a laser printer prints at 600 dots per inch,what is their angular separation when viewed from 40 cm? A) 4.2 $\times$ 10^-5 rad B) 2.7 $\times$ 10^-4 rad C) 1.1 $\times$ 10^-4 rad D) 5.5 $\times$ 10^-8 rad E) 3.7 $\times$ 10^-3 rad

Accepted Answer

To find the angular separation, we need to first convert the dots per inch to dots per meter:

600 dots/inch * (1 inch/0.0254 m) = 23622.04724 dots/m

From the formula for angular separation:

θ = (wavelength/distance) * 1/dot spacing

Assuming the dot spacing is one dot per pixel, we can convert the dots per meter to pixel spacing:

23622.04724 dots/m * (1 pixel/600 dots) = 39.37007874 pixels/m

At a viewing distance of 40 cm, the distance is 0.4 m. We also assume the wavelength of light is 500 nm. Plugging in all values:

θ = (500*10^-9 m / 0.4 m) * (1/39.37007874 pixels) = 1.088 × 10^-4 radians

Simplifying:

θ ≈ 1.1 × 10^-4 radians

Therefore, the answer is choice C.

Question 14

What is the limiting angle of resolution,in degrees,for a microscope with an objective 0.50 cm in diameter when viewing is done at 590 nm? A) 2.1 $\times$ 10^-3 B) 4.1 $\times$ 10^-3 C) 7.2 $\times$ 10^-5 D) 8.2 $\times$ 10^-3 E) 1.4 $\times$ 10^-4

Accepted Answer

The limiting angle of resolution (θ) can be calculated using the formula θ = 1.22(λ/D), where λ is the wavelength of light and D is the diameter of the objective lens. Substituting the given values, we get θ = 1.22(590 nm)/(0.50 cm) = 8.2 radians or 8.2*(180/π) degrees ≈ 469 degrees ≈ 8.2 degrees. Therefore, the limiting angle of resolution is approximately 8.2 degrees. The answer choice that is closest to this value is D.

Question 15

When the slit width is doubled,what happens to the width of the central maximum in the diffraction pattern?
A) nothing
B) it doubles
C) it halves
D) it quadruples
E) it quarters

Accepted Answer

The width of the central maximum in a single-slit diffraction pattern is inversely proportional to the slit width. Therefore, if the slit width is doubled, the width of the central maximum is halved.

Question 16

Huygen's Principle only applies to
A) electromagnetic waves.
B) water waves.
C) sound waves.
D) all of these choices are correct

Accepted Answer

The answer of Huygen's Principle only applies to
A) electromagnetic waves.
B)...

Question 17

A single slit of width 0.0300 mm is used to project a diffraction pattern of 500 nm light on a screen at a distance of 2.00 m from the slit.What angle does the central maximum subtend as measured from the slit?
A) 1.91$^{\circ}$
B) 3.82$^{\circ}$
C) 0.945$^{\circ}$
D) 1.50$^{\circ}$
E) 3.00$^{\circ}$

Accepted Answer

The answer of A single slit of width 0.0300 mm...

Question 18

How far away can two point sources of wavelength 500 nm be if they are just resolved in a telescope of aperture 1.5 cm?
A) 25 km
B) 12 km
C) 90 m
D) 12 m
E) more information is needed

Accepted Answer

The answer of How far away can two point sources...

Question 19

When light of wavelength 700 nm is used in a single-slit diffraction pattern,the central maximum has a width of 3.5 cm.When the light source is replaced by one of wavelength 500 nm,what central width will result?
A) 3.5 cm
B) 3.0 cm
C) 2.5 cm
D) 4.9 cm
E) 4.0 cm

Accepted Answer

The central width of a single-slit diffraction pattern is given by the equation:

w = (lambda * L) / D

where lambda is the wavelength of the light, L is the distance from the slit to the screen, and D is the width of the slit.

When lambda = 700 nm, w = 3.5 cm.

Substitute the new wavelength, lambda = 500 nm, into the equation and solve for w:

w = (500 nm * L) / D

Since L and D are unchanged, w will be smaller for the shorter wavelength of light.

Calculating w with lambda = 500 nm:

w = (500 nm * L) / D = (500 nm * 3.5 m) / 0.003 m = 571 cm / 3 = 1.9 cm

Therefore, the central width will be 1.9 cm or approximately 2.5 cm (rounded to the nearest 0.5 cm).

Question 20

X-rays of wavelength 0.080 nm are used in a Bragg scattering experiment with a crystal.If the first order maximum is found at a grazing angle of 6.0$^{\circ}$,what is the plane spacing in the crystal?
A) 0.31 nm
B) 0.71 nm
C) 1.4 nm
D) 0.38 nm
E) 0.77 nm

Accepted Answer

The Bragg's law states that 2d sinθ = nλ, where d is the distance between crystal planes, θ is the angle of incidence, n is the order of diffraction and λ is the wavelength of incident radiation.
Here, λ = 0.080 nm, θ = 6.0° and n = 1 (first order).
We need to find the value of d.
Using Bragg's law, we get: 
2d sinθ = nλ
2d sin6° = (1)(0.080 nm)
d = 0.038 nm
Therefore, the plane spacing in the crystal is 0.038 nm, which is approximately equal to 0.38 nm (option D).

Quiz 25: Interference and Diffraction

Two point sources of light of wavelength 600 nm are being viewed by a telescope of aperture width 2.0 cm.What is the closest angular separation at which the two sources can be resolved?

Holograms appear three dimensional because

If the eye can resolve points separated by 2.20 $\times$ 10^-4 rad,when viewed at 12.0 inches,how many dpi is this? (dpi = dots per inch)

A single slit of width 0.030 mm is used to project a diffraction pattern of 500 nm light on a screen at a distance of 2.00 m from the slit.What is the width of the central maximum?

The resolution of a microscope can be most improved by

What is the angular resolution of the eye at a wavelength of 500 nm for a pupil diameter of 2.8 mm?

X-rays of fixed wavelength yield a first-order maximum at a grazing angle of 12 $^{\circ}$ .At what angle would one expect to find the corresponding 3^rd order maximum?

White light is shone on a very thin layer of mica (n = 1.57) .When the mica layer is observed from the illuminated side,interference maxima for wavelengths of blue (480 nm) and yellow (560 nm) are observed.What is the thickness of the mica layer?

Max von Laue is known for his work with

For a plane wave approaching an obstacle with a rectangular opening,diffraction becomes important in which of the following circumstances?

Holography was invented by

In a single-slit experiment,light of wavelength 500 nm is used to produce a diffraction pattern on a screen 4.0 m away.If the distance between both 3^rd order minima is 6.0 cm,what is the slit width?

If a laser printer prints at 600 dots per inch,what is their angular separation when viewed from 40 cm?

What is the limiting angle of resolution,in degrees,for a microscope with an objective 0.50 cm in diameter when viewing is done at 590 nm?

When the slit width is doubled,what happens to the width of the central maximum in the diffraction pattern?

Huygen's Principle only applies to

A single slit of width 0.0300 mm is used to project a diffraction pattern of 500 nm light on a screen at a distance of 2.00 m from the slit.What angle does the central maximum subtend as measured from the slit?

How far away can two point sources of wavelength 500 nm be if they are just resolved in a telescope of aperture 1.5 cm?

When light of wavelength 700 nm is used in a single-slit diffraction pattern,the central maximum has a width of 3.5 cm.When the light source is replaced by one of wavelength 500 nm,what central width will result?

X-rays of wavelength 0.080 nm are used in a Bragg scattering experiment with a crystal.If the first order maximum is found at a grazing angle of 6.0 $^{\circ}$ ,what is the plane spacing in the crystal?