Question 1

Laser light of 589 nm wavelength is shone through parallel slits forming a series of maxima on a screen 4.20 m away. If the distance between the central maximum and the first-order maximum is 1.67 cm, what is the slit separation? A) 3.90 × 10^-3 cm B) 3.98 × 10^-3 m C) 9.98 × 10^-5 m D) 1.40 × 10^-3 m E) 1.48 × 10^-4 m

Accepted Answer

The answer of Laser light of 589 nm wavelength is...

Question 2

If light of wavelength 690 nm is used in a double-slit experiment with slit separation 0.206 mm, what angle separates the two fourth-order maxima?
A) 0.768°
B) 3.07°
C) 0.307°
D) 1.54°
E) 0.154°

Accepted Answer

The answer of If light of wavelength 690 nm is...

Question 3

Substituting tan θ for sin θ results in what percent error for an angle of 5.00°?
A) 5.00
B) 0.0166
C) 0.0333
D) 0.191
E) 0.382

Accepted Answer

The percent error can be calculated using the formula: $$ \text{Percent Error} = \left( \frac{\text{Approximate Value} - \text{True Value}}{\text{True Value}} \right) \times 100\% $$For an angle of 5.00°:- True Value = $\sin(5.00°)$- Approximate Value = $\tan(5.00°)$First, convert the angle to radians because the trigonometric functions in most calculators use radians. $5.00° = \frac{5 \times \pi}{180} \approx 0.0873$ radians.- $\sin(5.00°) \approx \sin(0.0873) \approx 0.0872$- $\tan(5.00°) \approx \tan(0.0873) \approx 0.0875$Now, calculate the percent error:$$ \text{Percent Error} = \left( \frac{0.0875 - 0.0872}{0.0872} \right) \times 100\% \approx 0.34\% $$However, none of the provided choices exactly match this calculation. The closest provided choice is 0.382, which suggests a possible miscalculation or approximation in the conversion or calculation steps. Given the options, the closest match and the intended answer is likely based on a more precise calculation or a different interpretation of the calculation steps.

Question 4

For an angle of 10 degrees, what is the percent difference between sin θ and θ in radians?
A) 0.1%
B) 0.3%
C) 0.5%
D) 0.7%
E) more than 1%

Accepted Answer

For small angles, we can use the approximation sin θ ≈ θ (in radians).
So, sin θ = sin(10°) ≈ 0.1745 and θ in radians = 10° × π/180 ≈ 0.1745.
The percent difference can be found using the formula: (|sin θ - θ|/((sin θ + θ)/2)) × 100%
Plugging in the values, we get:
(|0.1745 - 0.1745|/((0.1745 + 0.1745)/2)) × 100% = 0%
Therefore, the percent difference between sin θ and θ in radians is 0%, which is closest to option C: 0.5%.

Question 5

In a double-slit experiment, when both interference and diffraction are taken into account, which is the lowest order maximum that is missing in the case where the slit separation is 0.020 mm and the slit widths are each 0.005 mm?
A) 1
B) 2
C) 3
D) 4
E) no order is missing

Accepted Answer

The answer of In a double-slit experiment, when both interference...

Question 6

In a double-slit interference pattern, the third maximum from the central maximum corresponds to which value of m?
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The distance between the central maximum and the third maximum in a double-slit interference pattern is given by d sin θ = 3λ/2, where d is the distance between the slits, θ is the angle between the line from the center of the screen to the third maximum and a line perpendicular to the screen, and λ is the wavelength of light used in the experiment. The third maximum occurs when the path difference between the waves from the two slits is 3λ/2. Using the formula for path difference, we can write this as d sin θ = 3λ/2. Rearranging the formula, we get sinθ = (3λ/2)/d. Using the small angle approximation, we can write sinθ ≈ θ, so we get θ ≈ (3λ/2)/d. Using the formula for the position of the maxima in a double-slit interference pattern, we can write sinθ = mλ/d, where m is an integer that represents the order of the maximum (m = 0 for the central maximum, m = +/- 1 for the first order maxima, m = +/- 2 for the second order maxima, etc.). Equating the two expressions for sinθ, we get mλ/d ≈ (3λ/2)/d, which simplifies to m ≈ 3/2. Since m must be an integer, the only possible value is m = 2 for the third maximum. Therefore, the answer is B) 2.

Question 7

A Young's double-slit experiment is performed and then repeated after moving the screen to three times the previous distance from the slits. What happens to the angular separation (as measured from the slits) of the fourth-order maxima?
A) There is no change.
B) It becomes one-third the original value.
C) It becomes four times the original value.
D) It triples.
E) It increases by a factor of 2.25.

Accepted Answer

The angular separation of the maxima in a Young's double-slit experiment is given by $\theta = \frac{\lambda}{d}$, where $\lambda$ is the wavelength of light and $d$ is the slit separation. This formula shows that the angular separation is independent of the distance between the slits and the screen, meaning moving the screen does not change the angular separation of the maxima.

Question 8

A laser beam is directed through a grating and a first-order maximum is produced at 17°. What is the highest order maximum that can be produced by this arrangement?
A) 2
B) 3
C) 4
D) 5
E) 6

Accepted Answer

The highest order maximum, $m$, for a diffraction grating can be found using the grating equation $d\sin(\theta) = m\lambda$, where $d$ is the grating spacing, $\theta$ is the diffraction angle, and $\lambda$ is the wavelength of the light. Since the first-order maximum ($m=1$) occurs at 17°, for higher orders, the angle must increase. However, the sine function has a maximum value of 1, corresponding to an angle of 90°. Therefore, to find the highest possible order, we set $\sin(\theta) = 1$, which gives $d\sin(90°) = m\lambda$. Since $d\sin(17°) = \lambda$ for the first order, by increasing the angle to 90°, we are essentially finding how many times $\lambda$ can fit into $d\sin(90°)$ or equivalently, how many times the first-order condition can be met within the physical constraints of the sine function. Given that $\sin(17°)$ is significantly less than 1, and considering the discrete nature of $m$, the highest order achievable before exceeding the sine function's maximum is 3, as going to the fourth order would require a sine value greater than 1, which is impossible.

Question 9

Using a double slit with d = 300 nm and light of wavelength 400 nm, what would you see on a distant screen?
A) The intensity is maximum at the center and gradually decreases to either side.
B) The intensity is minimal at the center and gradually increases to either side.
C) The intensity is maximum in a single narrow line at the center.
D) The intensity is minimal in a single narrow line at the center.

Accepted Answer

According to the double-slit experiment, when light passes through two slits, it creates an interference pattern on a screen a distance away. The pattern will consist of alternating bright and dark fringes. The intensity of the fringes decreases as you move away from the center, with the brightest fringe being at the center. Therefore, the correct answer is A: The intensity is maximum at the center and gradually decreases to either side.

Question 10

If the two second-order maxima (m = ±2) are separated by 2.0 cm on the screen in a double-slit experiment, what is the separation of the m = ±3 maxima? Assume the angles are very small.
A) 1.5 cm
B) 2.0 cm
C) 2.5 cm
D) 3.0 cm
E) 3.5 cm

Accepted Answer

The separation between maxima in a double-slit experiment is directly proportional to the order of the maxima (m). If the separation for m = ±2 is 2.0 cm, then for m = ±3, which is 1.5 times greater, the separation would be 1.5 times 2.0 cm, resulting in 3.0 cm.

Question 11

For an angle of one degree, to how many significant figures do sin θ and θ (in radians) agree?
A) 3
B) 4
C) 5
D) 6
E) 7

Accepted Answer

One degree is approximately 0.0174533 radians. To four significant figures, this is 0.01745. sin θ can be approximated as θ for small angles, so sin θ to four significant figures is also 0.01745. Therefore, sin θ and θ agree to four significant figures.

Question 12

Light of wavelength 450 nm produces a first-order maximum at 27° when viewed through a grating. At what angle would a first-order maximum occur for a wavelength of 600 nm viewed through this grating?
A) 54°
B) 40°
C) 36°
D) 37°
E) 14°

Accepted Answer

We can use the formula for the grating equation, which is:
dsinθ = mλ
where d is the spacing of the grating (not given), θ is the angle at which the maximum occurs, m is the order of the maximum (1 in this case), and λ is the wavelength of light. 
For the first maximum at 450 nm, we have:
dsin(27°) = (1)(450 nm)
Now we can solve for d:
d = (450 nm) / sin(27°) = 996 nm
Then for the first maximum at 600 nm:
dsinθ = (1)(600 nm)
sinθ = (600 nm) / d
θ = sin⁻¹[(600 nm) / d] = sin⁻¹[(600 nm) / 996 nm] ≈ 37°
Therefore, the answer is D, 37°.

Question 13

Increasing the wavelength in a double-slit experiment has what effect on the position of the maxima on a screen at fixed distance?
A) none
B) maxima get closer together
C) maxima get farther apart
D) maxima get cancelled by minima
E) maxima pass minima on the screen

Accepted Answer

The position of the maxima on a screen at fixed distance is directly proportional to the wavelength of the light. As the wavelength increases, the distance between the maxima also increases. Therefore, the correct answer is C- maxima get farther apart.

Question 14

What is the highest order maximum possible using a grating with 600 slits per mm for a wavelength of 700 nm?
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The answer of What is the highest order maximum possible...

Question 15

A laser beam passes through a grating and produces a first-order maximum at an angle of 25°. At what angle is the second-order maximum?
A) 75°
B) 58°
C) 50°
D) 35°
E) 27°

Accepted Answer

For a diffraction grating, the angle $	heta$ of the order maximum is given by $d\sin(	heta) = m\lambda$, where $d$ is the spacing between the grating lines, $\lambda$ is the wavelength of light, and $m$ is the order of the maximum.

For the first-order maximum, we have $m=1$ and $	heta=25^\circ$. Let us assume that the wavelength of light is $\lambda$. Then,

$$d\sin(25^\circ) = \lambda$$

For the second-order maximum, we have $m=2$. Let us assume that the angle of this maximum is $	heta'$ and substitute this into the equation above. Then,

$$d\sin(	heta') = 2\lambda$$

Now, we can divide these two equations to eliminate $d$:

$$\frac{\sin(	heta')}{\sin(25^\circ)} = 2$$

Solving for $	heta'$ gives:

$$	heta' = \sin^{-1}(2\sin(25^\circ)) \approx 57.8^\circ$$

Therefore, the angle of the second-order maximum is approximately 58°, which corresponds to choice (B).

Question 16

What slit separation is required for a grating to produce a first-order maximum at an angle of 40.0° for a wavelength of 589 nm?
A) 109 nm
B) 690 nm
C) 378 nm
D) 458 nm
E) 916 nm

Accepted Answer

The grating equation is d*sin(θ) = m*λ, where d is the slit separation, θ is the angle, m is the order number, and λ is the wavelength. For the first-order maximum (m=1) at 40.0° with λ = 589 nm, solving for d gives d = λ/sin(θ) = 589 nm / sin(40°) ≈ 916 nm.

Question 17

A double-slit experiment is performed and then redone using slits of double the previous separation. Nothing else is changed. If the approximation sin θ ≈ θ holds, what happens to the angle between the maxima?
A) it stays the same
B) it doubles
C) it quadruples
D) it halves
E) it quarters

Accepted Answer

Doubling the slit separation in a double-slit experiment halves the angle between the maxima because the angle θ for the maxima is inversely proportional to the slit separation.

Question 18

In a double-slit experiment, the intensity of the light reaching the center of the screen from one slit alone is I0 and intensity of the light reaching the center from the other slit alone is 9I0. When both slits are open, what is the intensity of the light at the minima nearest the center? A) 4I₀ B) 0 C) 8I₀ D) 2I₀

Accepted Answer

The intensity at the minima in a double-slit experiment is given by $I_{min} = ( \sqrt{I_1} - \sqrt{I_2} )^2$, where $I_1$ and $I_2$ are the intensities from each slit. Here, $I_1 = I_0$ and $I_2 = 9I_0$, so $I_{min} = ( \sqrt{I_0} - \sqrt{9I_0} )^2 = (1I_0^{1/2} - 3I_0^{1/2})^2 = 4I_0$.

Question 19

In a Young's double-slit experiment, a fifth-order maximum occurs at an angle of 1.422°. If the screen is 3.90 m from the slits and the slit separation is 0.135 mm, what wavelength is being used?
A) 335 nm
B) 670 nm
C) 476 nm
D) 589 nm
E) 690 nm

Accepted Answer

The formula for the position of maxima in a double-slit experiment is $d \sin \theta = m \lambda$, where $d$ is the slit separation, $\theta$ is the angle, $m$ is the order of the maximum, and $\lambda$ is the wavelength. Substituting the given values: $0.135 \times 10^{-3} \sin(1.422^\circ) = 5 \lambda$. Solving for $\lambda$ gives approximately 670 nm.

Question 20

A grating is made with 600 slits per millimeter. What is the slit separation? A) 600 nm B) 1.67 × 10^-6 m C) 3.33 × 10^-6 m D) 160 pm E) 916 nm

Accepted Answer

The slit separation can be calculated using the formula d = 1/N, where N is the number of slits per unit length (in this case, per millimeter). Therefore, d = 1/600 mm = 1.67 × 10^-6 m, which is answer choice B.

Quiz 25: Interference and Diffraction

Laser light of 589 nm wavelength is shone through parallel slits forming a series of maxima on a screen 4.20 m away. If the distance between the central maximum and the first-order maximum is 1.67 cm, what is the slit separation?

If light of wavelength 690 nm is used in a double-slit experiment with slit separation 0.206 mm, what angle separates the two fourth-order maxima?

Substituting tan θ for sin θ results in what percent error for an angle of 5.00°?

For an angle of 10 degrees, what is the percent difference between sin θ and θ in radians?

In a double-slit experiment, when both interference and diffraction are taken into account, which is the lowest order maximum that is missing in the case where the slit separation is 0.020 mm and the slit widths are each 0.005 mm?

In a double-slit interference pattern, the third maximum from the central maximum corresponds to which value of m?

A Young's double-slit experiment is performed and then repeated after moving the screen to three times the previous distance from the slits. What happens to the angular separation (as measured from the slits) of the fourth-order maxima?

A laser beam is directed through a grating and a first-order maximum is produced at 17°. What is the highest order maximum that can be produced by this arrangement?

Using a double slit with d = 300 nm and light of wavelength 400 nm, what would you see on a distant screen?

If the two second-order maxima (m = ±2) are separated by 2.0 cm on the screen in a double-slit experiment, what is the separation of the m = ±3 maxima? Assume the angles are very small.

For an angle of one degree, to how many significant figures do sin θ and θ (in radians) agree?

Light of wavelength 450 nm produces a first-order maximum at 27° when viewed through a grating. At what angle would a first-order maximum occur for a wavelength of 600 nm viewed through this grating?

Increasing the wavelength in a double-slit experiment has what effect on the position of the maxima on a screen at fixed distance?

What is the highest order maximum possible using a grating with 600 slits per mm for a wavelength of 700 nm?

A laser beam passes through a grating and produces a first-order maximum at an angle of 25°. At what angle is the second-order maximum?

What slit separation is required for a grating to produce a first-order maximum at an angle of 40.0° for a wavelength of 589 nm?

A double-slit experiment is performed and then redone using slits of double the previous separation. Nothing else is changed. If the approximation sin θ ≈ θ holds, what happens to the angle between the maxima?

In a double-slit experiment, the intensity of the light reaching the center of the screen from one slit alone is I0 and intensity of the light reaching the center from the other slit alone is 9I0. When both slits are open, what is the intensity of the light at the minima nearest the center?

In a Young's double-slit experiment, a fifth-order maximum occurs at an angle of 1.422°. If the screen is 3.90 m from the slits and the slit separation is 0.135 mm, what wavelength is being used?

A grating is made with 600 slits per millimeter. What is the slit separation?