# Light from Element X through a diffraction Grating

• kikko
In summary, light emitted by Element X is passed through a diffraction grating with 1200 lines/mm. The diffraction pattern is observed on a screen 75.0 cm behind the grating, with bright fringes seen at 56.2 cm, 65.9 cm, and 93.5 cm from the central maximum. The maximum value of m for each of these diffracted wavelengths is unknown. The range of wavelengths for the visible spectrum is also unknown. To find the shortest wavelength, m values need to be determined until the first question can be answered. The equations used in this problem include the angles of bright fringes (\theta_m), the distance between fringes (y_m), the wavelength of light

## Homework Statement

Light emitted by Element X passes through a diffraction grating having 1200 lines/mm. The diffraction pattern is observed on a screen 75.0 cm behind the grating. Bright fringes are seen on the screen at distances of 56.2 cm, 65.9 cm, and 93.5 cm from the central maximum. No other fringes are seen.

What is the maximum value of m for each of these diffracted wavelengths?
This part will be visible after you complete previous part(s).
What are the wavelengths of light emitted by Element X?

## Homework Equations

$$\theta$$m = m($$\lambda/d$$) (angles of bright fringes)
ym = (m$$\lambda$$L)/d
dsin$$\theta$$m = m$$\lambda$$
ym - Ltan$$\theta$$m

## The Attempt at a Solution

\theta

No clue for this one. I tried plugging various things into the second equation, but got lost. I also don't understand how it can have multiple $$\lambda$$.

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What does each variable represent?

θm, m, λ, d, ym, L

kikko said:

## Homework Statement

Light emitted by Element X passes through a diffraction grating having 1200 lines/mm. The diffraction pattern is observed on a screen 75.0 cm behind the grating. Bright fringes are seen on the screen at distances of 56.2 cm, 65.9 cm, and 93.5 cm from the central maximum. No other fringes are seen.

What is the maximum value of m for each of these diffracted wavelengths?
This part will be visible after you complete previous part(s).
What are the wavelengths of light emitted by Element X?

## Homework Equations

$$\theta$$m = m($$\lambda/d$$) (angles of bright fringes)
ym = (m$$\lambda$$L)/d
dsin$$\theta$$m = m$$\lambda$$
ym - Ltan$$\theta$$m

## The Attempt at a Solution

\theta

No clue for this one. I tried plugging various things into the second equation, but got lost. I also don't understand how it can have multiple $$\lambda$$.

What is the range of wavelengths for the visible spectrum?

Find where the shortest wavelength (for the visible) would a appear on the screen, for m=1, m=2, ... until you can completely answer the first question.

I would approach this problem by first understanding the concepts of diffraction and diffraction gratings. A diffraction grating is a device with many parallel slits or lines that can diffract light into a spectrum of colors. In this case, we are given that the diffraction grating has 1200 lines/mm, which means that it has a high resolution and can produce a detailed diffraction pattern.

To determine the maximum value of m for each of the diffracted wavelengths, we can use the equation dsin\thetam = m\lambda, where d is the distance between the lines on the grating, \theta is the angle of diffraction, and m is the order of the diffraction. We are given that the bright fringes are seen at distances of 56.2 cm, 65.9 cm, and 93.5 cm from the central maximum. This means that m can have values of 1, 2, and 3 respectively. Therefore, the maximum value of m for each of the diffracted wavelengths is 3.

To calculate the wavelengths of light emitted by Element X, we can use the equation ym = (m\lambdaL)/d, where ym is the distance from the central maximum, L is the distance from the grating to the screen, and d is the distance between the lines on the grating. We are given that L = 75.0 cm and d = 1/1200 mm. Plugging in the values for m and ym, we can solve for \lambda. The calculated wavelengths are 560 nm, 659 nm, and 935 nm for m = 1, 2, and 3 respectively.

In summary, the diffraction pattern observed on the screen is a result of the diffraction of light from Element X passing through the diffraction grating. The maximum value of m for each of the diffracted wavelengths is 3, and the wavelengths of light emitted by Element X are 560 nm, 659 nm, and 935 nm.

## What is diffraction grating and how does it work?

Diffraction grating is a scientific instrument used to separate light into its component wavelengths. It works by utilizing a series of closely spaced parallel lines or grooves, which act as tiny slits that diffract the light passing through them.

## How does diffraction grating help us study light from element X?

By using a diffraction grating, we can split the light from element X into its individual wavelengths. This allows us to analyze and measure the specific wavelengths present in the light, providing valuable information about the element's composition and properties.

## What are the benefits of using a diffraction grating over other methods of studying light?

Unlike other methods, such as prisms, diffraction gratings produce a much clearer and more precise separation of light into its component wavelengths. This makes it a more accurate and reliable tool for studying and analyzing the properties of light from element X.

## How is the diffraction pattern of light from element X through a diffraction grating determined?

The diffraction pattern is determined by the spacing and number of lines on the grating. The distance between the lines affects the angle at which the light is diffracted, and the number of lines determines the intensity and shape of the diffraction pattern.

## What are some real-world applications of diffraction grating in studying light from element X?

Diffraction gratings are commonly used in spectroscopy, which is the study of the interaction between matter and electromagnetic radiation. They are also used in astronomy to analyze the light from stars and other celestial objects, providing valuable information about their composition and properties.