For a single-slit diffraction experiment, the condition for the first diffraction minimum is given by:
a sin θ = λ a \sin \theta = \lambda a sin θ = λ
where:
a a a θ \theta θ λ \lambda λ
To find the fractional error in the measurement of wavelength λ \lambda λ ln λ = ln a + ln ( sin θ ) \ln \lambda = \ln a + \ln(\sin \theta) ln λ = ln a + ln ( sin θ )
Differentiating both sides gives the relative change in λ \lambda λ d λ λ = d a a + cos θ d θ sin θ = d a a + cot θ d θ \frac{d\lambda}{\lambda} = \frac{da}{a} + \frac{\cos \theta \, d\theta}{\sin \theta} = \frac{da}{a} + \cot \theta \, d\theta λ d λ = a d a + s i n θ c o s θ d θ = a d a + cot θ d θ
Thus, the maximum fractional error in λ \lambda λ Δ λ λ = Δ a a + cot θ ⋅ Δ θ \frac{\Delta \lambda}{\lambda} = \frac{\Delta a}{a} + \cot \theta \cdot \Delta \theta λ Δ λ = a Δ a + cot θ ⋅ Δ θ
Step 1: Calculate the fractional error in slit width (Δ a a \frac{\Delta a}{a} a Δ a
Given:
a = 0.016 mm a = 0.016 \text{ mm} a = 0.016 mm Δ a = 0.002 mm \Delta a = 0.002 \text{ mm} Δ a = 0.002 mm
Δ a a = 0.002 mm 0.016 mm = 1 8 = 0.125 \frac{\Delta a}{a} = \frac{0.002 \text{ mm}}{0.016 \text{ mm}} = \frac{1}{8} = 0.125 a Δ a = 0.016 mm 0.002 mm = 8 1 = 0.125
Step 2: Calculate the term cot θ ⋅ Δ θ \cot \theta \cdot \Delta \theta cot θ ⋅ Δ θ
Given:
θ = 2 ∘ \theta = 2^\circ θ = 2 ∘ sin ( 2 ∘ ) = 0.035 \sin(2^\circ) = 0.035 sin ( 2 ∘ ) = 0.035 Δ θ = 40 ′ = ( 40 60 ) ∘ = ( 2 3 ) ∘ \Delta \theta = 40' = \left(\frac{40}{60}\right)^\circ = \left(\frac{2}{3}\right)^\circ Δ θ = 4 0 ′ = ( 60 40 ) ∘ = ( 3 2 ) ∘
Converting Δ θ \Delta \theta Δ θ Δ θ = 2 3 × π 180 rad = π 270 rad ≈ 0.011636 rad \Delta \theta = \frac{2}{3} \times \frac{\pi}{180} \text{ rad} = \frac{\pi}{270} \text{ rad} \approx 0.011636 \text{ rad} Δ θ = 3 2 × 180 π rad = 270 π rad ≈ 0.011636 rad
Since sin ( 2 ∘ ) = 0.035 \sin(2^\circ) = 0.035 sin ( 2 ∘ ) = 0.035 cos ( 2 ∘ ) = 1 − sin 2 ( 2 ∘ ) ≈ 1 \cos(2^\circ) = \sqrt{1 - \sin^2(2^\circ)} \approx 1 cos ( 2 ∘ ) = 1 − sin 2 ( 2 ∘ ) ≈ 1 cot ( 2 ∘ ) \cot(2^\circ) cot ( 2 ∘ ) cot ( 2 ∘ ) = cos ( 2 ∘ ) sin ( 2 ∘ ) ≈ 1 0.035 = 200 7 \cot(2^\circ) = \frac{\cos(2^\circ)}{\sin(2^\circ)} \approx \frac{1}{0.035} = \frac{200}{7} cot ( 2 ∘ ) = s i n ( 2 ∘ ) c o s ( 2 ∘ ) ≈ 0.035 1 = 7 200
Now, compute cot θ ⋅ Δ θ \cot \theta \cdot \Delta \theta cot θ ⋅ Δ θ cot ( 2 ∘ ) ⋅ Δ θ = 200 7 × π 270 = 20 π 189 ≈ 20 × 3.14159 189 ≈ 0.3324 \cot(2^\circ) \cdot \Delta \theta = \frac{200}{7} \times \frac{\pi}{270} = \frac{20\pi}{189} \approx \frac{20 \times 3.14159}{189} \approx 0.3324 cot ( 2 ∘ ) ⋅ Δ θ = 7 200 × 270 π = 189 20 π ≈ 189 20 × 3.14159 ≈ 0.3324
Step 3: Total Fractional Error
Combining both components:
Δ λ λ = 0.125 + 0.3324 = 0.4574 \frac{\Delta \lambda}{\lambda} = 0.125 + 0.3324 = 0.4574 λ Δ λ = 0.125 + 0.3324 = 0.4574
Rounding to two decimal places, the fractional error in the wavelength measurement is approximately 0.46 (or within the range 0.43 − 0.50 0.43 - 0.50 0.43 − 0.50