Question 1

Single-slit diffraction: A laser beam passes through a thin slit. When the pattern is viewed on a screen 1.25 m past the slit, you observe that the fifth-order dark fringes occur at ±2.41 cm from the central bright fringe. The entire experiment is now performed within a liquid, and you observe that each of the fifth-order dark fringes is 0.790 cm closer to the central fringe than it was in air. What is the index of refraction of this liquid?

Accepted Answer

The position of the dark fringes in air is given by:
$$x_n = n\frac{\lambda L}{d}$$
where $n$ is the order of the fringe, $\lambda$ is the wavelength of the laser, $L$ is the distance from the slit to the screen, and $d$ is the width of the slit.

For the fifth-order fringe, we have:
$$2x_5 = 2\cdot5\frac{\lambda L}{d} = 0.0241	ext{ m}$$
In the liquid, the wavelength of the laser is reduced by a factor of the refractive index $n$ of the liquid:
$$\lambda' = \frac{\lambda}{n}$$
The distance between fringes is still given by $x_n = n\frac{\lambda' L}{d}$. For the fifth-order fringe, we have:
$$2x_5' = 2\cdot5\frac{\lambda' L}{d} = 0.0169	ext{ m}$$
The difference in position between the fifth-order fringes in air and in the liquid is:
\begin{align*}
\Delta x_5 &= x_5'-x_5 \
&= 0.0241	ext{ m} - 0.0169	ext{ m} \
&= 0.0072	ext{ m}
\end{align*}
This is equal to the amount by which each fringe has moved closer to the central fringe:
$$\Delta x_5 = 0.790	ext{ cm} = 0.0079	ext{ m}$$
Therefore, we have:
$$\frac{\Delta x_5}{x_5} = \frac{0.0072	ext{ m}}{0.0241	ext{ m}} = \frac{0.0079	ext{ m}}{0.0241	ext{ m}-0.0079	ext{ m}} = n-1$$
Solving for $n$, we get:
$$n \approx 1.49$$
Therefore, the index of refraction of the liquid is approximately 1.49. The best choice is (C).

Question 2

Intensity of single-slit pattern: A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maximum is I₀, what is the intensity 10° from the center?

Accepted Answer

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Question 3

Diffraction grating: Monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. If the third-order maximum of the diffraction pattern is observed at 32.0° from the central spot, how many TOTAL number of maxima can be seen?

Accepted Answer

The formula for the maxima in a diffraction grating is given by $d\sin\theta = m\lambda$, where $d$ is the grating spacing, $\theta$ is the angle of the maxima from the central maximum, $m$ is the order of the maximum, and $\lambda$ is the wavelength of the light. Given that the third-order maximum ($m = 3$) is observed at 32.0°, we can use the equation to find the maximum possible value of $m$ that still produces a real value for $\sin\theta$, which must be less than or equal to 1. However, the key insight is that for each order $m$ on one side of the central maximum, there is a corresponding order $-m$ on the other side, plus the central maximum ($m = 0$). If the third-order maximum is visible, it implies that orders $m = -3, -2, -1, 0, 1, 2,$ and $3$ are visible, totaling 7 maxima. But since the question asks for the total number of maxima, we must consider that higher orders could also be visible if they do not exceed the angle at which light can still diffract. Given the angle for $m = 3$, we can infer that orders $m = 4$ and $m = -4$ are also possible before the sine of the angle exceeds 1. This gives us a total of $3 (for m = 1, 2, 3) + 3 (for m = -1, -2, -3) + 1 (for m = 0) + 2 (for m = 4 and m = -4) = 11$ maxima.

Question 4

Intensity of single-slit pattern: A single slit that is 2100 nm wide forms a diffraction pattern when illuminated by monochromatic light of 680-nm wavelength. At an angle of 10° from the central maximum, what is the ratio of the intensity to the intensity of the central maximum?

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Question 5

Single-slit diffraction: If the fifth order minimum in the diffraction pattern of a thin slit is at 40° from the central maximum, what is the highest order minimum in the pattern?

Accepted Answer

The condition for minima in single-slit diffraction is given by $ a \sin \theta = m \lambda $, where $ a $ is the slit width, $ \theta $ is the angle, $ m $ is the order of the minimum, and $ \lambda $ is the wavelength. The maximum possible value for $ \sin \theta $ is 1, so the highest order $ m $ is when $ m \lambda = a $. Given that the fifth order minimum is at 40°, we can calculate the highest order by considering the limit $ \sin \theta = 1 $. Solving for $ m $ gives the highest order as 7.

Question 6

Single-slit diffraction: A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the first two minima on either side of the central maximum?

Accepted Answer

The angular position of the minima in single-slit diffraction is given by the formula θ = mλ/a, where m is the order of the minimum, λ is the wavelength, and a is the slit width. For the first minima (m=1), θ = (1)(550 x 10^-9 m) / (0.050 x 10^-3 m) = 0.011 radians. Converting to degrees, θ ≈ 0.63°. The angular separation between the first minima on either side is 2θ, which is approximately 0.63°.

Question 7

Single-slit diffraction: If the fifth order minimum in the diffraction pattern due to a thin slit is at 40° from the central maximum, at what angle does the first order minimum occur?

Accepted Answer

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Question 8

Single-slit diffraction: A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen. The 5^th minimum occurs at 7.00° away from the central maximum. At what angle does the 18^th minimum occur?

Accepted Answer

The angle for the nth minimum in a single-slit diffraction pattern is given by $n\lambda = d\sin(\theta)$, where $n$ is the order of the minimum, $\lambda$ is the wavelength of the light, $d$ is the slit width, and $\theta$ is the angle of the minimum from the central maximum. The angle for the nth minimum is directly proportional to $n$. Given that the first minimum occurs at 7.00°, the angle for the 18th minimum can be found by scaling this angle by 18, since the angles are directly proportional to the order of the minima. Therefore, the angle for the 18th minimum is $18 \times 7.00° = 126.0°$. However, this calculation is incorrect based on the initial premise; the correct approach involves understanding that the angle for the nth minimum is not simply $n$ times the angle of the first minimum due to the sine function relationship. The correct answer must be determined by understanding the relationship between the order of the minimum, the wavelength, and the slit width, and applying this to the specific question of the 18th minimum, which involves more complex calculations or understanding of diffraction patterns beyond the simple linear scaling implied by the incorrect explanation. Given the options and the nature of diffraction patterns, the correct angle must be one that is feasible within the context of single-slit diffraction, where higher order minima occur at larger angles, but the relationship is not linear as implied. The correct answer, therefore, involves recognizing the mistake in the initial explanation and understanding the principles of diffraction, which would typically result in an angle that is a realistic outcome of the diffraction formula for the 18th minimum, considering the provided options and the context of the question.

Question 9

Intensity of single-slit pattern: A thin beam of light of wavelength 625 nm goes through a thin slit and falls on a screen 3.00 m past the slit. You observe that the first completely dark fringes occur on the screen at distances of ±8.24 mm from the central bright fringe, and that the central bright fringe has an intensity of 2.00 W/m2 at its center.
(a) How wide is the slit?
(b) What is the intensity of light at a point on the screen that is one-quarter of the way from the central bright fringe to the first dark fringe?

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Question 10

Intensity of single-slit pattern: If the central maximum in a single slit diffraction pattern has intensity I₀, the intensity of the secondary maximum that is nearest to it is closest to

Accepted Answer

The intensity of the first secondary maximum in a single-slit diffraction pattern is about 4.5% (or 0.045 times) of the intensity of the central maximum, making option A correct.

Question 11

Single-slit diffraction: A slit of width 0.010 mm has light of frequency 5.0 × 10¹⁴ Hz passing through it onto a screen 60 cm away. How wide is the central maximum? (c = 3.00 × 10⁸ m/s)

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Question 12

Single-slit diffraction: In the figure, a slit 0.30 mm wide is illuminated by light of wavelength 426 nm. A diffraction pattern is seen on a screen 2.8 m from the slit. What is the linear distance on the screen between the first diffraction minima on either side of the central diffraction maximum?

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Question 13

Single-slit diffraction: Light of wavelength 687 nm is incident on a single slit 0.75 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the diffraction pattern?

Accepted Answer

The position of the dark fringes in a single-slit diffraction pattern is given by the formula $y = \frac{m\lambda L}{a}$, where $y$ is the distance from the center of the pattern to the fringe on the screen, $m$ is the order of the fringe (for the second dark fringe, $m = 2$), $\lambda$ is the wavelength of the light (687 nm = $687 \times 10^{-9}$ m), $L$ is the distance from the slit to the screen, and $a$ is the width of the slit (0.75 mm = $0.75 \times 10^{-3}$ m). Rearranging for $L$, we get $L = \frac{ya}{m\lambda}$.Substituting the given values, $L = \frac{1.7 \times 10^{-3} \times 0.75 \times 10^{-3}}{2 \times 687 \times 10^{-9}}$, we find $L \approx 0.93$ meters, which corresponds to choice C.

Question 14

Single-slit diffraction: A single slit forms a diffraction pattern with monochromatic light. The 4th minimum of the pattern occurs at an angle of 35° from the central maximum. How many bright bands are on each side of the central maximum?

Accepted Answer

The number of bright bands on each side of the central maximum is one less than the number of minima on that side. Since the 4th minimum occurs at 35°, there are 3 bright bands on each side of the central maximum, making a total of 4 bright bands including the central maximum.

Question 15

Diffraction grating: A diffraction grating has 300 lines per mm. If light of wavelength 630 nm is sent through this grating, what is the highest order maximum that will appear?

Accepted Answer

The formula to find the highest order maximum is mλ = d(sinθ). Here, m is the order maximum, λ is the wavelength of light, d is the distance between adjacent grooves on the grating, and θ is the angle between the incident light and the direction of the maximum. Solving for m, we get m = (dsinθ)/λ. The highest order maximum occurs when sinθ = 1, which gives m = d/λ. Substituting the given values, we get m = (1/300)/630x10^-9 = 5. So the highest order maximum that will appear is 5.

Question 16

Diffraction grating: When monochromatic light illuminates a grating with 7000 lines per centimeter, its second order maximum is at 62.4°. What is the wavelength of the light?

Accepted Answer

The formula to find the wavelength of light is:
d * sinθ = mλ
Where d is the spacing between the grating lines, θ is the angle at which the maximum occurs (in radians), m is the order of the maximum, and λ is the wavelength of the light.

In this case, d = 1/7000 cm (since there are 7000 lines per 1 cm), θ = 62.4° = 1.09 radians (converted from degrees), and m = 2 (since we are looking for the second order maximum).

Plugging these values into the formula, we get:

(1/7000 cm) * sin(1.09 radians) = 2λ

Solving for λ, we get:

λ = (1/2) * (1/7000 cm) * sin(1.09 radians) = 6.33 x 10^-5 cm = 633 nm

Question 17

Diffraction grating: A diffraction grating has 300 lines per mm. If light of frequency 4.76 × 10¹⁴ Hz is sent through this grating, at what angle does the first order maximum occur? (c = 3.00 × 10⁸ m/s)

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Question 18

Diffraction grating: In a diffraction grating experiment, light of 600 nm wavelength produces a first-order maximum 0.350 mm from the central maximum on a distant screen. A second monochromatic source produces a third-order maximum 0.870 mm from the central maximum when it passes through the same diffraction grating. What is the wavelength of the light from the second source?

Accepted Answer

For the first source:
d*sin(theta) = m*lambda
where d is the slit separation, theta is the angle of diffraction, m is the order of diffraction and lambda is the wavelength.
At central maximum, sin(theta) = 0
So, for the first-order maximum,
0.35*10^-3 = (1)(600*10^-9)d
d = 0.5833*10^-3 m
For the second source:
At the same angle of diffraction theta, the path difference between the slits for the second source will be 2*2*d = 4d (as the order of diffraction is 3).
So, for the third-order maximum,
0.87*10^-3 = (3)(lambda)*d/(4d)
lambda = 497*10^-9 m = 497 nm.

Question 19

Diffraction grating: Monochromatic light is incident on a grating that is 75 mm wide and ruled with 50,000 lines. The second-order maximum is seen at 32.5°. What is the wavelength of the incident light?

Accepted Answer

The wavelength ($\lambda$) can be found using the diffraction grating equation $\lambda = \frac{d \sin \theta}{m}$, where $d$ is the grating spacing (the inverse of the number of lines per meter), $\theta$ is the angle of the diffraction order, and $m$ is the order number. Given 50,000 lines per 75 mm, or 50,000 lines per 0.075 m, the spacing $d = \frac{1}{50000/0.075} = \frac{0.075}{50000}$ meters. For the second-order maximum ($m = 2$) at 32.5°, $\lambda = \frac{(0.075/50000) \sin(32.5°)}{2}$. Calculating gives a wavelength close to 403 nm.

Question 20

Diffraction grating: A He-Ne laser, which produces light of wavelength 632.8 nm, is used to calibrate a diffraction grating. If the first-order maximum occurs at 20.5° from the central spot, what is the distance between the slits of the grating?

Accepted Answer

The distance between the slits of the grating can be calculated using the formula:

d * sin θ = mλ

where d is the distance between the slits, θ is the angle of diffraction, m is the order of the maximum, and λ is the wavelength of the light.

In this case, we have:

d * sin 20.5° = (1) * 632.8 nm

Solving for d, we get:

d = (1 * 632.8 nm) / sin 20.5°

d ≈ 1.81 µm

Therefore, the correct answer is B.

Quiz 28: Sources of Magnetic Field

Intensity of single-slit pattern: A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maximum is I0, what is the intensity 10° from the center?