Explaining 2.2 x 10⁻⁶ m Range Calculation

[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

 

[Steve4Physics]

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?

 

[Charles Link]

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.

 

[Charles Link]

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]

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.

 

[Steve4Physics]

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.

I’m saying answer C) is the answer to a different question:

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.

I’m saying answer C) is the answer to a different question:

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?

 

[Steve4Physics]

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/.