# Homework Help: Calculating the Number of Lines on a Diffraction Grating

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1. Jun 19, 2016

### EmilyBergendahl

1. The problem statement, all variables and given/known data
A diffraction grating gives a second-order maximum at as angle of 31° for violet light (λ = 4.0 × 10^2 nm). If the diffraction grating is 1.0 cm in width, how many lines are on this diffraction grating?

2. Relevant equations
d = (m)(λ)/sinθm

3. The attempt at a solution
d = (m)(λ)/sinθm
d = (2)(4.0 x 10-7 m)/sin(31°)
d = 1.6 x 10 x 10-6 m

0.01 m / 1.6 1.6 x 10 x 10-6 m = 6250 lines

Many thanks in advance!

Last edited: Jun 19, 2016
2. Jun 19, 2016

You need to use the equation $m \lambda=d \sin{\theta}$ for the location $\theta$ of an interference maximum. From this equation and the info they gave you, you can compute "d"=the distance between the lines on the grating. (The grating is filled with these lines. Item of interest=your computed "d" is going to be very small=on the order of a wavelength of light.) The width of the grating is w=1.0 cm so you need quite a large number N of these closely spaced lines to make 1.0 cm. Do you know what the letter "m" represents?

3. Jun 19, 2016

### EmilyBergendahl

Is this what you mean? I was in the process of editing as you posted.

4. Jun 19, 2016

Yes. Your d=1.6 E-6 (meters) is what it should read. I believe your answer is correct N=6250 but I would need to doublecheck the arithmetic. (And your m=2 is correct.)

5. Jun 19, 2016

### EmilyBergendahl

Yes, whoops, got a little overzealous with the scientific notation there, haha.

Thank you Charles!

6. Mar 27, 2018

### Zeynep Celik

why is m = 2?

7. Mar 27, 2018

In the statement of the problem in the OP (original post=post 1), it states that it is a "second order maximum" that they are observing. Thereby $m=2$ in the equation $m \lambda=d \sin{\theta}$.

8. Mar 27, 2018

### Zeynep Celik

didn't notice that, thanks!

9. Apr 12, 2018

### Vignesh Mallaiah

What is highest order spectrum which may be seen with monochromatic light of wavelength 600nm by means of a diffraction grating with 5000 lines/cm

I need answer for this question pls help me.