A diffraction grating has \(4.00 \cdot 10^{3}\) lines \(/ \mathrm{cm}\) and has white light \((400 .-700 . \mathrm{nm})\) incident on it. What wavelength(s) will be visible at \(45.0^{\circ} ?\)

Short Answer

Expert verified
Answer: No visible wavelengths will appear at the given angle of 45.0° for the diffraction grating with 4.00 x 10³ lines/cm.

Step by step solution

01

Calculate the grating spacing

To find the grating spacing (d), we need to divide 1 cm by the number of lines per centimeter. In this case, the grating has 4.00 x 10³ lines/cm, so: \(d = \frac{1}{4.00 \times 10^3 \text{lines/cm}}\) \(d = 2.50 \times 10^{-4} \mathrm{cm}\)
02

Convert grating spacing to meters

Since we need to calculate wavelengths in nanometers, it is better to convert the grating spacing into meters: \(d = 2.50 \times 10^{-4} \mathrm{cm} \times \frac{1 \mathrm{m}}{100 \mathrm{cm}}\) \(d = 2.50 \times 10^{-6} \mathrm{m}\)
03

Use the diffraction grating formula to find the order

The formula for the diffraction grating is given by: \(dsin(\theta) = m\lambda\) Where \(d\) is the grating spacing, \(\theta\) is the angle of diffraction, \(m\) is the order of the diffraction, and \(\lambda\) is the wavelength of the light. In our case, \(\theta = 45.0°\), and we need to find the wavelength, so we can rearrange the formula: \(\lambda = \frac{dsin(\theta)}{m}\) \(m = \frac{dsin(\theta)}{\lambda}\)
04

Substitute grating spacing and angle to find the order

Now we substitute the grating spacing (\(d = 2.50 \times 10^{-6} \mathrm{m}\)) and the angle (\(\theta = 45.0°\)) into the formula and find the order: \(m = \frac{2.50 \times 10^{-6} \mathrm{m} \cdot sin(45.0°)}{\lambda}\)
05

Determine the wavelengths for each order

Using the minimum and maximum values of the white light, we find the ranges of orders for each wavelength: For \(\lambda_{min} = 400 \mathrm{nm}\): \(m_{min} = \frac{2.50 \times 10^{-6} \mathrm{m} \cdot sin(45.0°)}{400 \times 10^{-9} \mathrm{m}} \approx 2.48\) For \(\lambda_{max} = 700 \mathrm{nm}\): \(m_{max} = \frac{2.50 \times 10^{-6} \mathrm{m} \cdot sin(45.0°)}{700 \times 10^{-9} \mathrm{m}} \approx 1.45\) Since \(m_{min}\) and \(m_{max}\) lie between 1 and 2, we can only use the range of wavelengths for the first order \(m = 1\) to find the visible wavelengths at \(45.0°\).
06

Calculate the visible wavelengths

Now we can use the formula to find the visible wavelengths for the first-order diffraction (\(m = 1\)): \(\lambda = \frac{2.50 \times 10^{-6} \mathrm{m} \cdot sin(45.0°)}{1}\) \(\lambda \approx 1.77 \times 10^{-6} \mathrm{m} = 1770 \mathrm{nm}\) Since 1770 nm falls outside the range of white light (\(400 - 700 \mathrm{nm}\)), there are no visible wavelengths for the first-order diffraction at \(45.0^{\circ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Grating Spacing
Understanding the concept of grating spacing is crucial when studying diffraction gratings. Grating spacing, denoted by the symbol d, refers to the distance between adjacent slits or lines in a diffraction grating. It is inversely related to the number of lines per unit length. In essence, if you have a grating with a high number of lines per centimeter, the spacing between these lines will be very small, and vice versa.

When approaching a problem, the first step is often to calculate this grating spacing since it plays a significant role in determining the diffraction pattern. In our exercise, the grating spacing is found by taking the reciprocal of the number of lines per centimeter. Mathematically, we express this as:
\[d = \frac{1}{\text{number of lines/cm}}\]
For a grating with 4,000 lines per centimeter, the spacing is 2.50 x 10-4 cm. Converting to meters is necessary as it provides consistency when using the SI system for subsequent calculations, particularly in the formula where wavelengths are typically expressed in meters or nanometers.
Wavelength Calculation
Wavelength calculation is at the heart of understanding how a diffraction grating interacts with light. Given that different wavelengths of light will diffract at different angles, calculating the wavelength can be essential for predicting the color or intensity of light in various diffraction orders. For this, we use the diffraction grating equation:
\[d\sin(\theta) = m\lambda\]
This formula correlates the grating spacing d, the angle of diffraction θ, the order of diffraction m, and the wavelength of light λ. To find any one of these variables, we rearrange the equation accordingly.

In the given problem, we needed to calculate the resulting wavelength(s) of light at a specific angle (45.0°). For this, we isolate λ from the equation:
\[\lambda = \frac{d\sin(\theta)}{m}\]
However, in this case, the calculated wavelength falls outside the visible spectrum, indicating that at the angle given, no visible light is observed. This demonstrates how precise measuring and calculations are integral in wavelength determination and also highlights the importance of range—understanding which wavelengths correspond to visible light (typically 400 - 700 nm).
Diffraction Orders
The term 'diffraction orders' refers to the series of spectra produced by the diffraction grating, numbered according to the integer m in the grating equation. These orders represent different instances where constructive interference occurs for different wavelengths. The first order (m=1) is the first series of diffracted light, the second order (m=2) is the next, and so on. Each of these orders will appear at different angles for the same wavelength of light.

In our exercise, we observed that the possible orders for visible light at 45.0° fell between the first and second order, which means only the first order could be viable for white light's given range (400 - 700 nm). However, after calculating the exact wavelength that would appear at this angle for the first order, we found it to be outside this range. This leads us to conclude that, while the concept of orders is fundamental to understanding diffraction patterns, the visibility of these orders is subject to the restrictions of the visible light spectrum and the particular conditions of the diffraction setup, such as the angle of observation.

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Most popular questions from this chapter

Think of the pupil of your eye as a circular aperture \(5.00 \mathrm{~mm}\) in diameter. Assume you are viewing light of wavelength \(550 \mathrm{nm}\), to which your eyes are maximally sensitive. a) What is the minimum angular separation at which you can distinguish two stars? b) What is the maximum distance at which you can distinguish the two headlights of a car mounted \(1.50 \mathrm{~m}\) apart?

A double slit is opposite the center of a 1.8 -m wide screen \(2.0 \mathrm{~m}\) from the slits. The slit separation is \(24 \mu \mathrm{m}\) and the width of each slit is \(7.2 \mu \mathrm{m}\). How many fringes are visible on the screen if the slit is illuminated by \(600 .-\mathrm{nm}\) light?

At the Long-baseline Interferometer Gravitationalwave Observatory (LIGO) facilities in Hanford, Washington, and Livingston, Louisiana, laser beams of wavelength \(550.0 \mathrm{nm}\) travel along perpendicular paths \(4.000 \mathrm{~km}\) long. Each beam is reflected along its path and back 100 times before the beams are combined and compared. If a gravitational wave increases the length of one path and decreases the other, each by 1.000 part in \(10^{21}\), what is the phase difference between the two beams as a result?

A diffraction grating with exactly 1000 lines per centimeter is illuminated by a He-Ne laser of wavelength \(633 \mathrm{nm}\). a) What is the highest order of diffraction that could be observed with this grating? b) What would be the highest order if there were exactly 10,000 lines per centimeter?

Calculate and compare the angular resolutions of the Hubble Space Telescope (aperture diameter \(2.40 \mathrm{~m}\), wavelength \(450 . \mathrm{nm}\); illustrated in the text), the Keck Telescope (aperture diameter \(10.0 \mathrm{~m}\), wavelength \(450 . \mathrm{nm}\) ), and the Arecibo radio telescope (aperture diameter \(305 \mathrm{~m}\), wavelength \(0.210 \mathrm{~m}\) ). Assume that the resolution of each instrument is diffraction limited.

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