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Wave Optics question

2025 · Shift 1 · Q44
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Wave Optics question

2025 · Shift 1 · Q44

JEE AdvancedPhysicsWave OpticsNumerical+4 / −1
A single slit diffraction experiment is performed to determine the slit width using the equation, bdD=mλ\frac{b d}{D}=m \lambdaDbd​=mλ, where bbb is the slit width, DDD the shortest distance between the slit and the screen, ddd the distance between the mth m^{\text {th }}mth  diffraction maximum and the central maximum, and λ\lambdaλ is the wavelength. DDD and ddd are measured with scales of least count of 1 cm and 1 mm , respectively. The values of λ\lambdaλ and mmm are known precisely to be 600 nm and 3, respectively. The absolute error (in μm\mu \mathrm{m}μm) in the value of bbb estimated using the diffraction maximum that occurs for m=3m=3m=3 with d=5 mmd=5 \mathrm{~mm}d=5 mm and D=1 mD=1 \mathrm{~m}D=1 m is ‾\underline{\hspace{2cm}}​.
Numerical answer
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