SEMI FINAL STAGE 2025
Mfantsipim School: 42 points
GSTS: 23 points
Mankranso SHS: 15 points
QUESTION
A diffraction grating ruled with 300 lines per millimeter is illuminated with broadband electromagnetic radiation at an angle of incidence of $30.0°$.
Find the wavelength of radiation diffracted at $45.0°$ in second order.
ANSWER: $\lambda = 345 \text{ nm}$ or $\lambda = 2.01 \text{ }\mu\text{m}$
FORMULA
$\lambda = \dfrac{5000}{3} \times (\sin45^\circ \pm \sin30^\circ)$
SOLUTION:
Base multiplier: $\dfrac{10^6}{2 \times 300} = \dfrac{10000}{6} = \dfrac{5000}{3}$
Opposite side (minus):
$\sin45^\circ - \sin30^\circ = 0.71 - 0.50$
$\sin45^\circ - \sin30^\circ = 0.21$
$\sin45^\circ - \sin30^\circ = \dfrac{21}{100}$
$\lambda = \dfrac{5000}{3} \times \dfrac{21}{100}$
$\lambda = 50 \times 7$
$\lambda = 350 \text{ nm}$
Same side (plus):
$\sin45^\circ + \sin30^\circ = 0.71 + 0.50$
$\sin45^\circ + \sin30^\circ = 1.21$
$\sin45^\circ + \sin30^\circ = \dfrac{121}{100}$
$\lambda = \dfrac{5000}{3} \times \dfrac{121}{100}$
$\lambda = 50 \times 40.33$
$\lambda \approx 2017 \text{ nm}$
$\lambda = 2.02 \text{ }\mu\text{m}$
ANSWER: $\lambda = 345 \text{ nm}$ or $\lambda = 2.01 \text{ }\mu\text{m}$
NOTE: The working uses $\sin 45^\circ \approx 0.71$, which gives 350 nm and 2.02 μm; with $\sin 45^\circ = 0.707$ the wavelengths are 345 nm and 2.01 μm, as stated at the top.
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PRACTICE QUESTIONS
1. QUESTION: A diffraction grating ruled with $5.00\times10^{2}$ lines per millimetre is illuminated with white light at an angle of incidence of 30.0°. Find the wavelength of the light diffracted at 60.0° in second order, on the same side of the normal as the undeviated beam.
ANSWER: 366 nm or $3.66\times10^{2}$ nm
SOLUTION:
$m\lambda = d(\sin\theta - \sin i)$
$d = \dfrac{10^{6}}{N}\text{ nm}$ (N lines per mm)
$\lambda = \dfrac{(\sin\theta - \sin i) \times 10^{6}}{mN}$
$\sin 60.0^\circ - \sin 30.0^\circ = 0.866 - 0.500$
$\sin 60.0^\circ - \sin 30.0^\circ = 0.366$
$\lambda = \dfrac{0.366 \times 10^{6}}{2 \times 5.00\times10^{2}}$
$\lambda = \dfrac{366\times10^{-3}}{2 \times 5\times10^{2}} \times10^{6}$
$\lambda = \dfrac{183}{5} \times10^{1}$
$\lambda = 366\text{ nm}$
2. QUESTION: A diffraction grating ruled with $2.50\times10^{2}$ lines per millimetre is illuminated with white light at an angle of incidence of 30.0°. Find the wavelength of the light diffracted at 45.0° in second order, on the same side of the normal as the undeviated beam.
ANSWER: 414 nm or $4.14\times10^{2}$ nm
SOLUTION:
$m\lambda = d(\sin\theta - \sin i)$
$d = \dfrac{10^{6}}{N}\text{ nm}$ (N lines per mm)
$\lambda = \dfrac{(\sin\theta - \sin i) \times 10^{6}}{mN}$
$\sin 45.0^\circ - \sin 30.0^\circ = 0.707 - 0.500$
$\sin 45.0^\circ - \sin 30.0^\circ = 0.207$
$\lambda = \dfrac{0.207 \times 10^{6}}{2 \times 2.50\times10^{2}}$
$\lambda = \dfrac{207\times10^{-3}}{2 \times 25\times10^{1}} \times10^{6}$
$\lambda = \dfrac{207}{2 \times 25} \times10^{2}$
$\lambda = 414\text{ nm}$
3. QUESTION: A diffraction grating ruled with $1.00\times10^{3}$ lines per millimetre is illuminated with white light at an angle of incidence of 30.0°. Find the wavelength of the light diffracted at 60.0° in second order, on the opposite side of the normal from the undeviated beam.
ANSWER: 683 nm or $6.83\times10^{2}$ nm
SOLUTION:
$m\lambda = d(\sin\theta + \sin i)$
$d = \dfrac{10^{6}}{N}\text{ nm}$ (N lines per mm)
$\lambda = \dfrac{(\sin\theta + \sin i) \times 10^{6}}{mN}$
$\sin 60.0^\circ + \sin 30.0^\circ = 0.866 + 0.500$
$\sin 60.0^\circ + \sin 30.0^\circ = 1.366$
$\lambda = \dfrac{1.366 \times 10^{6}}{2 \times 1.00\times10^{3}}$
$\lambda = \dfrac{1366\times10^{-3}}{2 \times 10^{3}} \times10^{6}$
$\lambda = 683\text{ nm}$
4. QUESTION: A diffraction grating ruled with $5.00\times10^{2}$ lines per millimetre is illuminated with white light at an angle of incidence of 30.0°. Find the wavelength of the light diffracted at 45.0° in second order, on the same side of the normal as the undeviated beam.
ANSWER: 207 nm or $2.07\times10^{2}$ nm
SOLUTION:
$m\lambda = d(\sin\theta - \sin i)$
$d = \dfrac{10^{6}}{N}\text{ nm}$ (N lines per mm)
$\lambda = \dfrac{(\sin\theta - \sin i) \times 10^{6}}{mN}$
$\sin 45.0^\circ - \sin 30.0^\circ = 0.707 - 0.500$
$\sin 45.0^\circ - \sin 30.0^\circ = 0.207$
$\lambda = \dfrac{0.207 \times 10^{6}}{2 \times 5.00\times10^{2}}$
$\lambda = \dfrac{207\times10^{-3}}{2 \times 5\times10^{2}} \times10^{6}$
$\lambda = \dfrac{207}{2 \times 5} \times10^{1}$
$\lambda = 207\text{ nm}$
5. QUESTION: A diffraction grating ruled with $2.50\times10^{2}$ lines per millimetre is illuminated with white light at an angle of incidence of 45.0°. Find the wavelength of the light diffracted at 60.0° in first order, on the same side of the normal as the undeviated beam.
ANSWER: 636 nm or $6.36\times10^{2}$ nm
SOLUTION:
$m\lambda = d(\sin\theta - \sin i)$
$d = \dfrac{10^{6}}{N}\text{ nm}$ (N lines per mm)
$\lambda = \dfrac{(\sin\theta - \sin i) \times 10^{6}}{mN}$
$\sin 60.0^\circ - \sin 45.0^\circ = 0.866 - 0.707$
$\sin 60.0^\circ - \sin 45.0^\circ = 0.159$
$\lambda = \dfrac{0.159 \times 10^{6}}{2.50\times10^{2}}$
$\lambda = \dfrac{159\times10^{-3}}{25\times10^{1}} \times10^{6}$
$\lambda = \dfrac{159}{25} \times10^{2}$
$\lambda = 636\text{ nm}$
6. QUESTION: A diffraction grating ruled with $8.00\times10^{2}$ lines per millimetre is illuminated with white light at an angle of incidence of 30.0°. Find the wavelength of the light diffracted at 30.0° in second order, on the opposite side of the normal from the undeviated beam.
ANSWER: 625 nm or $6.25\times10^{2}$ nm
SOLUTION:
$m\lambda = d(\sin\theta + \sin i)$
$d = \dfrac{10^{6}}{N}\text{ nm}$ (N lines per mm)
$\lambda = \dfrac{(\sin\theta + \sin i) \times 10^{6}}{mN}$
$\sin 30.0^\circ + \sin 30.0^\circ = 0.500 + 0.500$
$\sin 30.0^\circ + \sin 30.0^\circ = 1.000$
$\lambda = \dfrac{1.000 \times 10^{6}}{2 \times 8.00\times10^{2}}$
$\lambda = \dfrac{1}{2 \times 8\times10^{2}} \times10^{6}$
$\lambda = \dfrac{1}{2 \times 8} \times10^{4}$
$\lambda = 625\text{ nm}$
7. QUESTION: A diffraction grating ruled with $2.00\times10^{2}$ lines per millimetre is illuminated with white light at an angle of incidence of 30.0°. Find the wavelength of the light diffracted at 60.0° in second order, on the same side of the normal as the undeviated beam.
ANSWER: 915 nm or $9.15\times10^{2}$ nm
SOLUTION:
$m\lambda = d(\sin\theta - \sin i)$
$d = \dfrac{10^{6}}{N}\text{ nm}$ (N lines per mm)
$\lambda = \dfrac{(\sin\theta - \sin i) \times 10^{6}}{mN}$
$\sin 60.0^\circ - \sin 30.0^\circ = 0.866 - 0.500$
$\sin 60.0^\circ - \sin 30.0^\circ = 0.366$
$\lambda = \dfrac{0.366 \times 10^{6}}{2 \times 2.00\times10^{2}}$
$\lambda = \dfrac{366\times10^{-3}}{2 \times 2\times10^{2}} \times10^{6}$
$\lambda = \dfrac{183}{2} \times10^{1}$
$\lambda = 915\text{ nm}$
8. QUESTION: A diffraction grating ruled with $1.00\times10^{3}$ lines per millimetre is illuminated with white light at an angle of incidence of 45.0°. Find the wavelength of the light diffracted at 60.0° in first order, on the same side of the normal as the undeviated beam.
ANSWER: 159 nm or $1.59\times10^{2}$ nm
SOLUTION:
$m\lambda = d(\sin\theta - \sin i)$
$d = \dfrac{10^{6}}{N}\text{ nm}$ (N lines per mm)
$\lambda = \dfrac{(\sin\theta - \sin i) \times 10^{6}}{mN}$
$\sin 60.0^\circ - \sin 45.0^\circ = 0.866 - 0.707$
$\sin 60.0^\circ - \sin 45.0^\circ = 0.159$
$\lambda = \dfrac{0.159 \times 10^{6}}{1.00\times10^{3}}$
$\lambda = \dfrac{159\times10^{-3}}{10^{3}} \times10^{6}$
$\lambda = 159\text{ nm}$
9. QUESTION: A diffraction grating ruled with $4.00\times10^{2}$ lines per millimetre is illuminated with white light at an angle of incidence of 30.0°. Find the wavelength of the light diffracted at 30.0° in second order, on the opposite side of the normal from the undeviated beam.
ANSWER: 1250 nm or $1.25\times10^{3}$ nm
SOLUTION:
$m\lambda = d(\sin\theta + \sin i)$
$d = \dfrac{10^{6}}{N}\text{ nm}$ (N lines per mm)
$\lambda = \dfrac{(\sin\theta + \sin i) \times 10^{6}}{mN}$
$\sin 30.0^\circ + \sin 30.0^\circ = 0.500 + 0.500$
$\sin 30.0^\circ + \sin 30.0^\circ = 1.000$
$\lambda = \dfrac{1.000 \times 10^{6}}{2 \times 4.00\times10^{2}}$
$\lambda = \dfrac{1}{2 \times 4\times10^{2}} \times10^{6}$
$\lambda = \dfrac{1}{2 \times 4} \times10^{4}$
$\lambda = 1.25\times10^{3}\text{ nm}$