JEE Challenger
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PhyMedAdv 2025 • Paper 1

Absolute Error in Slit Width Determination from Single Slit Diffraction

A single slit diffraction experiment is performed to determine the slit width using the equation, bdD=mλ\frac{bd}{D} = m\lambda, where bb is the slit width, DD the shortest distance between the slit and the screen, dd the distance between the mthm^{\text{th}} diffraction maximum and the central maximum, and λ\lambda is the wavelength. DD and dd are measured with scales of least count of 1 cm1\text{ cm} and 1 mm1\text{ mm}, respectively. The values of λ\lambda and mm are known precisely to be 600 nm600\text{ nm} and 33, respectively. The absolute error (in μm\mu\text{m}) in the value of bb estimated using the diffraction maximum that occurs for m=3m = 3 with d=5 mmd = 5\text{ mm} and D=1 mD = 1\text{ m} is ____

Official Numerical Answer75 to 95

Topics & Concepts

Units and MeasurementsMeasuring InstrumentsDiffraction

Step-by-Step Solution

To find the absolute error in the slit width bb, we start from the given relation: b=mλDdb = \frac{m\lambda D}{d}

Given Data:

  • Order of diffraction maximum, m=3m = 3
  • Wavelength of light, λ=600 nm=600×10−9 m=0.6 μm\lambda = 600\text{ nm} = 600 \times 10^{-9}\text{ m} = 0.6\text{ }\mu\text{m}
  • Distance to the screen, D=1 m=100 cmD = 1\text{ m} = 100\text{ cm}
  • Least count (uncertainty) in DD, ΔD=1 cm=0.01 m\Delta D = 1\text{ cm} = 0.01\text{ m}
  • Position of the maximum, d=5 mmd = 5\text{ mm}
  • Least count (uncertainty) in dd, Δd=1 mm\Delta d = 1\text{ mm}

Step 1: Calculate the nominal value of bb

b=3×(600×10−9 m)×1 m5×10−3 m=360×10−6 m=360 μmb = \frac{3 \times (600 \times 10^{-9}\text{ m}) \times 1\text{ m}}{5 \times 10^{-3}\text{ m}} = 360 \times 10^{-6}\text{ m} = 360\text{ }\mu\text{m}

Step 2: Determine the absolute error Δb\Delta b

Method 1: Using the differential (fractional error) method

The relative error is given by: Δbb=ΔDD+Δdd\frac{\Delta b}{b} = \frac{\Delta D}{D} + \frac{\Delta d}{d}

Substitute the fractional errors: ΔDD=1 cm100 cm=0.01\frac{\Delta D}{D} = \frac{1\text{ cm}}{100\text{ cm}} = 0.01 Δdd=1 mm5 mm=0.20\frac{\Delta d}{d} = \frac{1\text{ mm}}{5\text{ mm}} = 0.20

Thus, Δbb=0.01+0.20=0.21\frac{\Delta b}{b} = 0.01 + 0.20 = 0.21

The absolute error in bb is: Δb=0.21×b=0.21×360 μm=75.6 μm\Delta b = 0.21 \times b = 0.21 \times 360\text{ }\mu\text{m} = 75.6\text{ }\mu\text{m}

Method 2: Using the finite maximum error method

The maximum possible estimated value of bb is: bmax⁡=mλ(D+ΔD)d−Δd=3×(600×10−9 m)×1.01 m4×10−3 m=454.5 μmb_{\max} = \frac{m\lambda (D + \Delta D)}{d - \Delta d} = \frac{3 \times (600 \times 10^{-9}\text{ m}) \times 1.01\text{ m}}{4 \times 10^{-3}\text{ m}} = 454.5\text{ }\mu\text{m}

The absolute error is: Δb=bmax⁡−b=454.5 μm−360 μm=94.5 μm\Delta b = b_{\max} - b = 454.5\text{ }\mu\text{m} - 360\text{ }\mu\text{m} = 94.5\text{ }\mu\text{m}

Final Answer:

Depending on the error formulation used:

  • By fractional error: 75.675.6 (Acceptable range: 7575 to 7979)
  • By finite maximum error: 94.594.5 (Acceptable range: 9494 to 9595)
Absolute Error in Slit Width Determination from Single Slit Diffraction | Physics PYQ Solution - JEE Challenger