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When yellow sodium light, \(\lambda=589 nm\), falls on a diffraction grating, its first-order peak on a screen 72.0 cm away falls 3.32 cm from the central peak. Another source produces a line 3.71 cm from the central peak.What is its wavelength? How many slits/cm are on the grating?

Short Answer

Expert verified
The unknown wavelength is approximately 658 nm, and the grating has about 725 slits/cm.

Step by step solution

01

Understand the Problem

We are given two sources of light falling on a diffraction grating: a yellow sodium light with a wavelength of \( \lambda_1 = 589 \) nm and another unknown wavelength \( \lambda_2 \). For the sodium light, the distance from the central peak is 3.32 cm for the first-order diffraction. For the unknown light, the distance is 3.71 cm. We need to find \( \lambda_2 \) and the grating density (number of slits/cm).
02

Apply Diffraction Grating Formula

The diffraction grating formula is \( d \sin \theta = m \lambda \), where \( d \) is the distance between the slits in the grating, \( \theta \) is the diffraction angle, \( m \) is the order of the spectrum, and \( \lambda \) is the wavelength. Since \( \theta \) is small, we can approximate \( \sin \theta \approx \tan \theta \approx \frac{y}{L} \), where \( y \) is the distance from the central peak and \( L \) is the distance to the screen.
03

Calculate Angle and Distance Between Slits for Sodium Light

For the sodium light, \( y_1 = 3.32 \) cm and \( L = 72.0 \) cm. Thus, \( \tan \theta_1 \approx \frac{3.32}{72} \approx 0.0461 \). Now, use \( d \tan \theta_1 = m \lambda_1 \). For \( m=1 \), \( d \times 0.0461 = 589 \) nm. Convert \( 589 \) nm to cm (\( 589 \ nm = 5.89 \times 10^{-5} \) cm) and solve for \( d \): \( d = \frac{5.89 \times 10^{-5}}{0.0461} \) cm.
04

Calculate Wavelength for Unknown Source

For the second light, \( y_2 = 3.71 \) cm. Calculate \( \tan \theta_2 \approx \frac{3.71}{72} \). Using the value of \( d \) found in Step 3, apply \( d \tan \theta_2 = \lambda_2 \): \( \lambda_2 = d \times \tan \theta_2 \). Substitute \( d \) and \( \tan \theta_2 \) to find \( \lambda_2 \).
05

Calculate Number of Slits per cm

From Step 3, you have \( d \). The number of slits per cm (grating density, \( N \)) is the reciprocal of \( d \). Therefore, \( N = \frac{1}{d} \). Calculate \( N \) using the value of \( d \) obtained previously.

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Key Concepts

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

Wavelength Calculation
When working with diffraction gratings, calculating the wavelength of a light source becomes crucial. The basic formula used here is derived from the diffraction grating equation.Consider light from a source of an unknown wavelength creating a diffracted pattern when passing through a diffraction grating. This pattern typically displays bright and dark fringes on an observing screen. The key steps to calculate the wavelength are:
  • Measure the distance from the central peak (zero-order maximum) to the first-order maximum on the screen. This is the value of \( y \).
  • Use the given distance between the screen and the grating, often denoted as \( L \).
  • Approximate \( \tan \theta = \frac{y}{L} \), which for small angles, equates to the sine.
The relationship \( d \tan \theta = m \lambda \) is used, where \( \lambda \) is the wavelength and \( m \) represents the order of the pattern. By solving these equations, the wavelength of the unknown light source can be determined.
Optical Physics
Optical physics involves studying light and its interactions with matter. Diffraction is a core phenomenon within this field. When light encounters an obstacle or slit, especially one comparable in size to its wavelength, it bends around it. This bending is called diffraction. A diffraction grating, which is a tool used in optical physics, helps in isolating light into its component wavelengths. It consists of closely spaced slits, allowing only certain light wavelengths to interfere constructively. This principles help scientists identify and analyze light sources with precision. The angle at which light emerges from a grating depends on its wavelength, thus providing a powerful analytical tool. Optical physics utilizes these principles in numerous applications, from creating spectroscopy equipment in laboratories to developing technologies like holography.
Grating Density
Grating density refers to the number of slits per centimeter on a diffraction grating. It is a fundamental characteristic that defines how a grating diffracts light. The higher the grating density, the closer the slits, thus allowing finer resolution in separating component wavelengths.To find the grating density, first calculate the slit spacing \( d \). If \( d \) is the distance between slits, then the grating density \( N \) is simply the reciprocal of \( d \):\[ N = \frac{1}{d} \]This means, if the distance between slits is known, one can easily calculate how many slits are fitted into a centimeter. Understanding grating density is crucial, as it influences the precision of wavelength measurement, and determines the spread of diffraction patterns. When designing scientific instruments, choosing the right grating density is essential for obtaining the desired resolution.

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Most popular questions from this chapter

Unpolarized light falls on two polarizer sheets whose axes are at right angles. (a) What fraction of the incident light intensity is transmitted? (b) What fraction is transmitted if a third polarizer is placed between the first two so that its axis makes a 56\(^\circ\) angle with the axis of the first polarizer? (c) What if the third polarizer is in front of the other two?

Light of wavelength \(5.0\times 10{-7}\) passes through two parallel slits and falls on a screen 5.0 m away. Adjacent bright bands of the interference pattern are 2.0 cm apart. (a) Find the distance between the slits. (b) The same two slits are next illuminated by light of a different wavelength, and the fifth-order minimum for this light occurs at the same point on the screen as the fourth-order minimum for the previous light.What is the wavelength of the second source of light?

(II) How many slits per centimeter does a grating have if the third order occurs at a 15.0\(^\circ\) angle for 620-nm light?

(I) Monochromatic light falling on two slits 0.018 mm apart produces the fifth-order bright fringe at an 8.6\(^\circ\) angle.What is the wavelength of the light used?

(III) Four polarizers are placed in succession with their axes vertical, at 30.0\(^\circ\) to the vertical, at 60.0\(^\circ\)to the vertical, and at 90.0\(^\circ\)to the vertical. (a) Calculate what fraction of the incident unpolarized light is transmitted by the four polarizers. (b) Can the transmitted light be decreased by removing one of the polarizers? If so, which one? (c) Can the transmitted light intensity be extinguished by removing polarizers? If so, which one(s)?

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