AQA A-level Physics Diffraction Gratings Mark Scheme

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Mark schemes
A
1 [1]

C
2 [1]

A
3 [1]

B
4 [1]

sinθ = nλ/d in this form/correct calculations of d/d = 1/300
5
C1

substitutes correctly – condone powers of 10

C1

18.9

C1

2 or 3 sf only

A1
[4]




(a) 1.06 × 10-6 m
6
B1
1




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(b) use of nλ = dsinθ with n = 2 and d = ans to (a)

C1

λ = 1.06 × 10-6 × sin 55º/2

C1

4.34 × 10-7 m

A1
3
[4]




(a) (i) = 590 × 10–9m (1)
7
(using d sin θ = nλ gives)



(1) = 0.707 or


7.07 × 108 if nm used (1)

θ = 45.0° (1) (accept 45°)



(ii) (sin θ ≤ 1) gives ≤ 1 or n ≤ or = (1) = 2.83 (1)

so 3rd order or higher order is not possible (1)

alternative solution:
(substituting) n = 3 (into d sin θ = nλ gives) (1)



sin θ ( ) = 1.06 (1)

gives ‘error’/which is not possible (1)
7

(b) (using d sin θ = nλ gives)

2 λ = 1.67 × 10–6 × sin 42.1 (1)

λ(= 0.5 × 1.67 × 10–6 × sin 42.1) = 5.6(0) × 10–7 m (or 560 nm) (1)
2
[9]

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(i) use of (26 ± 2) mm for 2θ or (13 ± 1) mm for θ
8
M1

range 2.60 → 3.02

A1

(ii) nλ= d sin θ seen

M1

substitution with correct powers
1.20 × 10−5 → 1.40 × 10−5

A1

1/d or number of lines m−1 calculated
71400 → 833300 needs unit

B1

conversion to number of lines mm−1
71(.4) → 83(.3) needs unit

B1
[6]




(a) d sin θ = n λ or d sin 12 = 6.3 x 10–7
9
C1

3.0 x 10–6 m

A1
2




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