# Light is incident normally on a diffraction grating, the ruling separation of the lines:

hidemi
Homework Statement:
If 550-nm light is incident normally on a diffraction grating and exactly 6 lines are produced, the ruling separation must be:

The answer is between 2.20 ´ 10-6 m and 3.30 ´ 10-6 m
Relevant Equations:
d·sin(θ) = n·λ
I get one of the ranges by calculating:

sin(90°) = 1
d·sin(θ) = d × 1 = d = n·λ
d = 6 × 550-nm = 3,300 nm = 3.3 microns = 3.3 × 10⁻⁶ m

But, how can I get the other range of 2.2 × 10⁻⁶ m

Could anyone explain it to me? Thanks

Homework Helper
Gold Member
2022 Award
Homework Statement:: If 550-nm light is incident normally on a diffraction grating and exactly 6 lines are produced, the ruling separation must be:
There can’t be ‘exactly 6 lines’ because there is a single 0-th order central line and pairs of higher order lines (symmetrical about centre). So there is an odd number of lines in total. For example, if the highest order line is 6 (n=6), there are exactly 2*6+1 = 13 lines visible.

Have you stated the question correctly?

Homework Helper
Gold Member
In general, the number of primary maxima will be odd, because of the ## m=0 ##, with ## m=\pm 1 ##, and ## m=\pm 2 ##, etc. I think the question needs additional clarification. Perhaps someone else has an idea of what it is referring to.

hidemi
There can’t be ‘exactly 6 lines’ because there is a single 0-th order central line and pairs of higher order lines (symmetrical about centre). So there is an odd number of lines in total. For example, if the highest order line is 6 (n=6), there are exactly 2*6+1 = 13 lines visible.

Have you stated the question correctly?
There isn't any typo. The picture of the original question is as attached.

#### Attachments

• 1.jpeg
35.6 KB · Views: 93
Homework Helper
Gold Member
I tried to solve it assuming there are orders up to and including ## m=5 ##. I can also see how they got the 3.3 but not the 2.2. I think they need to do their homework more carefully. The question is flawed.

Steve4Physics
Homework Helper
Gold Member
2022 Award
There isn't any typo. The picture of the original question is as attached.
The question is wrong (as already noted by @Charles Link and myself). And the so-called correct answer can't be correct because the limits of the ruling-separation range are 4λ and 6λ, which makes no sense.

I believe the question should be:

Light of wavelngth 550nm is incident normally on a diffraction grating. The highest order lines visible are the 6th-order. What is the possible range of the ruling-separations?

To answer this we consider the 2 limiting conditions.

a)The ruling-separation gives the 6th order line at θ = 89.9999999999º (but call it 90º!) which is just visible.

b)The ruling-separation gives the 7th order line at θ = 90º (you can’t actually see a line if θ = 90º). Note that the 6th order lines would now be visible at some angle smaller than 90º.

Using the amended question I get the answer C) in your list.

From what text-book (or other source) is the question?

hidemi
The question is wrong (as already noted by @Charles Link and myself). And the so-called correct answer can't be correct because the limits of the ruling-separation range are 4λ and 6λ, which makes no sense.

I believe the question should be:

Light of wavelngth 550nm is incident normally on a diffraction grating. The highest order lines visible are the 6th-order. What is the possible range of the ruling-separations?

To answer this we consider the 2 limiting conditions.

a)The ruling-separation gives the 6th order line at θ = 89.9999999999º (but call it 90º!) which is just visible.

b)The ruling-separation gives the 7th order line at θ = 90º (you can’t actually see a line if θ = 90º). Note that the 6th order lines would now be visible at some angle smaller than 90º.

Using the amended question I get the answer C) in your list.

From what text-book (or other source) is the question?
It's from class, and I'm not sure where the professor extracted from.
One of the answer C is above the limit of 3.3e-6, so I wonder if I misunderstood what you meant.

Homework Helper
Gold Member
2022 Award
One of the answer C is above the limit of 3.3e-6, so I wonder if I misunderstood what you meant.
There is no upper limit of 3.3e-6 m. Remember the original question is completel wrong - it is nonsense.

Light of wavelength 550nm is incident normally on a diffraction grating. The highest order lines visible are the 6th-order. What is the possible range of the ruling-separations?

##n\lambda= dsin\thetaθ## gives ##d = \frac{n\lambda}{sin\theta}##

If the 6th order is just visible, that means ##\theta = 89.99999999º##.
This corresponds to a line-spacing ##d = \frac {6\lambda}{sin(89.99999999º)} = 6\lambda##.

If the 7th order is just invisible that means ##\theta = 90º##.
This corresponds to a line-spacing ##d = \frac {7\lambda}{sin(90º)} = 7\lambda##.
(In this case the 6th order would easily be visible – you can work out, it is at 59º.)

Therefore the line-spacing (d) must be between ##6\lambda## and ##7\lambda##.

##d = 6 \times 550\times 10^{-9} m = 3.30\times 10^-6 m##.
##d = 7 \times 550\times 10^{-9} m = 3.85\times 10^-6 m##.

hidemi
There is no upper limit of 3.3e-6 m. Remember the original question is completel wrong - it is nonsense.

Light of wavelength 550nm is incident normally on a diffraction grating. The highest order lines visible are the 6th-order. What is the possible range of the ruling-separations?

##n\lambda= dsin\thetaθ## gives ##d = \frac{n\lambda}{sin\theta}##

If the 6th order is just visible, that means ##\theta = 89.99999999º##.
This corresponds to a line-spacing ##d = \frac {6\lambda}{sin(89.99999999º)} = 6\lambda##.

If the 7th order is just invisible that means ##\theta = 90º##.
This corresponds to a line-spacing ##d = \frac {7\lambda}{sin(90º)} = 7\lambda##.
(In this case the 6th order would easily be visible – you can work out, it is at 59º.)

Therefore the line-spacing (d) must be between ##6\lambda## and ##7\lambda##.

##d = 6 \times 550\times 10^{-9} m = 3.30\times 10^-6 m##.
##d = 7 \times 550\times 10^{-9} m = 3.85\times 10^-6 m##.
Thanks for the response.
I wonder how about the orders lie in between 0 < theta < 90?

Homework Helper
Gold Member
2022 Award
I wonder how about the orders lie in between 0 < theta < 90?
I don't understand the question. What you can try is this:

Pick a value for d.
Work out the angle for each order using
##\theta = sin^{-1}(\frac {n\lambda }{d})## for n=0, 1, 2, 3, ...

You will then see exactly where each order lies in the range 0≤θ<90º.

You can repeat this for a different value of d and see what difference it makes,

(If you can use a spreadsheet, this is very quick/easy to do.)

hidemi
I
I don't understand the question. What you can try is this:

Pick a value for d.
Work out the angle for each order using
##\theta = sin^{-1}(\frac {n\lambda }{d})## for n=0, 1, 2, 3, ...

You will then see exactly where each order lies in the range 0≤θ<90º.

You can repeat this for a different value of d and see what difference it makes,

(If you can use a spreadsheet, this is very quick/easy to do.)
I see, Thank you/.