# Calculating the Number of Lines on a Diffraction Grating

## Homework Statement

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?

d = (m)(λ)/sinθm

## 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:

Homework Helper
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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?

• EmilyBergendahl

## 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

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

Homework Helper
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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.)

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

Thank you Charles! • why is m = 2?

Homework Helper
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2020 Award
why is m = 2?
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} ##.

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} ##.
didn't notice that, thanks!

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