How Do You Calculate Diffraction Angles and Grating Lines?

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To calculate diffraction angles, use the formula d(sinθ) = mλ, where d is the slit separation, θ is the angle, m is the order of the maximum, and λ is the wavelength. For a pair of slits separated by 0.274 mm illuminated by 545.5 nm light, the angle to the first bright fringe is approximately 0.5031 degrees. In the case of a diffraction grating with a first-order maximum at 26.09 degrees for a 546.1 nm line, the number of lines per centimeter is calculated to be 3224. For a grating with 3224 slits/cm and a wavelength of 463.200 nm, the angle for the first-order spectral line is approximately 0.0048 degrees. Understanding these calculations is essential for analyzing diffraction patterns and spectral lines.
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Problem 3. A pair of narrow parallel slits separated by a distance of 0.274 mm are illuminated by the green component from a mercury vapor lamp (wavelength=545.5nm).
What is the angle from the central maximum to the first bright fringe on wither side of the central maximum? Answer in degrees.
Note: If the formula is: d(sin thetha)=m(wavelength)
What do i do next??

Problem 25.
A diffraction grating is calibrated by using the 546.1 m line of mercury vapor. The first-order maximum is found at an angle of 26.09 degrees. Calculate the number of lines per centimeter on this grating. Answer in units of lines/cm.
Note: How do I start?

Problem 12. By attaching a diffraction-grating spectroscope to an astronomical telescope, one can measure the spectral lines from a start and determine the start's chemical composition. Assume the grating has 3224 slits/cm. The wavelengths of the star's light are wavelength_1=463.200nm, wavelength_2=640.500 nm, and wavelength_3=704.700 nm.
Find the angle at which the first-order spectral line for wavelength_1 occurs. Answer in degrees.
Note: How do I start?
 
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Answer:Problem 3: To solve this problem, you will need to use the formula d(sinθ) = mλ, where d is the slit separation (given to be 0.274 mm), θ is the angle from the central maximum, m is an integer (in this case, m = 1), and λ is the wavelength of the light source (given to be 545.5 nm). Plugging in the given values, you get 0.274 sinθ = 545.5, so θ = 0.5031 degrees. This is the angle from the central maximum to the first bright fringe on either side of the central maximum.Problem 25: To solve this problem, you will need to use the formula mλ = dsinθ, where d is the slit separation, θ is the angle of the first-order maximum (given to be 26.09 degrees), m is an integer (in this case, m = 1), and λ is the wavelength of the light source (given to be 546.1 nm). Plugging in the given values, you get 546.1 = dsin26.09, so d = 3224 lines/cm. This is the number of lines per centimeter on the grating.Problem 12: To solve this problem, you will need to use the formula mλ = dsinθ, where d is the slit separation (given to be 3224 slits/cm), θ is the angle at which the first-order spectral line occurs, m is an integer (in this case, m = 1), and λ is the wavelength of the light (given to be 463.2 nm). Plugging in the given values, you get 463.2 = 3224sinθ, so θ = 0.0048 degrees. This is the angle at which the first-order spectral line for wavelength_1 occurs.
 


For Problem 3, you can use the formula d(sin theta) = m(wavelength) to solve for the angle theta. In this case, d is the distance between the slits (0.274 mm), m is the order of the bright fringe (1 for the first bright fringe), and lambda is the wavelength of green light (545.5 nm). So, the equation becomes 0.274 mm x (sin theta) = 1 x 545.5 nm. You can then solve for theta by dividing both sides by 0.274 mm and taking the inverse sine of the resulting value. This will give you the angle from the central maximum to the first bright fringe on either side.

For Problem 25, you can use the formula d(sin theta) = m(lambda) to solve for the number of lines per centimeter on the grating. In this case, d is the distance between adjacent lines on the grating (which we need to find), m is the order of the maximum (which is 1 for the first-order maximum), and lambda is the wavelength of the mercury vapor line (546.1 nm). So, the equation becomes d x (sin 26.09 degrees) = 1 x 546.1 nm. You can then solve for d by dividing both sides by (sin 26.09 degrees) and converting the wavelength to centimeters. This will give you the number of lines per centimeter on the grating.

For Problem 12, you can use the same formula d(sin theta) = m(lambda) to solve for the angle at which the first-order spectral line for wavelength_1 occurs. In this case, d is the distance between adjacent lines on the grating (which is given as 3224 slits/cm), m is the order of the maximum (which is 1 for the first-order maximum), and lambda is the wavelength of the first spectral line (463.200 nm). So, the equation becomes 3224 slits/cm x (sin theta) = 1 x 463.200 nm. You can then solve for theta by dividing both sides by 3224 slits/cm and taking the inverse sine of the resulting value. This will give you the angle at which the first-order spectral line for wavelength_1 occurs.
 
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