The Diffraction Grating | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

The Diffraction Grating

A diffraction grating has many equally spaced slits. Their waves add strongly only at particular angles, producing narrow principal maxima.

The central maximum is order

n=0

. Matching first, second and higher orders appear symmetrically on both sides.

Read the orders before calculating

A laser beam passing through a diffraction grating and splitting into five beams that land as five red dots on a screen — One laser beam in, a fan of sharp beams out: one bright dot for each order, brightest at n = 0.

The order number is counted from the centre. The first bright spot on either side is $n=1$ , not zero.

Use the grating relation

d\sin\theta=n\lambda

Symbols

$d$ = grating spacing: distance between adjacent slits (m)
$θ$ = angle from the undeviated central beam (°)
$n$ = order number: 0, 1, 2, ... ()
$λ$ = wavelength (m)

If the grating is labelled with a line density, invert it first:

d=\dfrac{1}{\text{lines per metre}}

Symbols

$d$ = grating spacing (m)

Common mistake

Convert the line density before inverting.

450\ \text{lines mm}^{-1}=450\,000\ \text{lines m}^{-1}

. Also measure

\theta

from the central beam, not from the other-side maximum.

Worked example

From line density to wavelength

A grating has 300 lines per mm. Its first-order maximum is at 10°. Find the wavelength.

Line density is $3.00\times10^5\ \text{m}^{-1}$ , so $d=3.33\times10^{-6}\ \text{m}$ .
With $n=1$ , $\lambda=d\sin10^\circ$ .

Answer

\lambda=5.8\times10^{-7}\ \text{m}

A possible order must satisfy $n\lambda\le d$ because $\sin\theta$ cannot exceed 1.
For the same grating, longer wavelengths appear at larger angles.
More lines per metre means a smaller spacing $d$ , so the same wavelength also appears at a larger angle.

Your turn9702/23/M/J/25 Q6(b)(ii)

The angle between the two second-order maxima for 720 nm light is 52°. Calculate the number of grating lines per metre. [3] (9702/23/M/J/25 Q6(b)(ii))

Show worked answer

The angle from the centre to one maximum is 26°.
$d=2(720\times10^{-9})/\sin26^\circ=3.3\times10^{-6}\ \text{m}$ .
Line density is $1/d$ .

Answer

3.0\times10^5\ \text{lines m}^{-1}

The Diffraction Grating

A diffraction grating has many equally spaced slits. Their waves add strongly only at particular angles, producing narrow principal maxima.

The central maximum is order

n=0

. Matching first, second and higher orders appear symmetrically on both sides.

Read the orders before calculating

The order number is counted from the centre. The first bright spot on either side is $n=1$ , not zero.

Use the grating relation

d\sin\theta=n\lambda

Symbols

$d$ = grating spacing: distance between adjacent slits (m)
$θ$ = angle from the undeviated central beam (°)
$n$ = order number: 0, 1, 2, ... ()
$λ$ = wavelength (m)

If the grating is labelled with a line density, invert it first:

d=\dfrac{1}{\text{lines per metre}}

Symbols

$d$ = grating spacing (m)

Common mistake

Convert the line density before inverting.

450\ \text{lines mm}^{-1}=450\,000\ \text{lines m}^{-1}

. Also measure

\theta

from the central beam, not from the other-side maximum.

Worked example

From line density to wavelength

A grating has 300 lines per mm. Its first-order maximum is at 10°. Find the wavelength.

Line density is $3.00\times10^5\ \text{m}^{-1}$ , so $d=3.33\times10^{-6}\ \text{m}$ .
With $n=1$ , $\lambda=d\sin10^\circ$ .

Answer

\lambda=5.8\times10^{-7}\ \text{m}

A possible order must satisfy $n\lambda\le d$ because $\sin\theta$ cannot exceed 1.
For the same grating, longer wavelengths appear at larger angles.
More lines per metre means a smaller spacing $d$ , so the same wavelength also appears at a larger angle.

Your turn9702/23/M/J/25 Q6(b)(ii)

The angle between the two second-order maxima for 720 nm light is 52°. Calculate the number of grating lines per metre. [3] (9702/23/M/J/25 Q6(b)(ii))

Show worked answer

The angle from the centre to one maximum is 26°.
$d=2(720\times10^{-9})/\sin26^\circ=3.3\times10^{-6}\ \text{m}$ .
Line density is $1/d$ .

Answer

3.0\times10^5\ \text{lines m}^{-1}