A single slit diffraction experiment is performed to determine the slit width using the equation, where is the slit width, D the shortest distance between the slit and the screen, the distance between the diffraction maximum and the central maximum, a

Question

A single slit diffraction experiment is performed to determine the slit width using the equation, $\frac{b d}{D} = m λ,$ where $b$ is the slit width, D the shortest distance between the slit and the screen, $d$ the distance between the $m^{t h}$ diffraction maximum and the central maximum, and $λ$ is the wavelength. D and $d$ are measured with scales of least count of 1 cm and 1 mm, respectively. The values of $λ$ and $m$ are known precisely to be 600 nm and 3, respectively. The absolute error (in $μ$ m) in the value of $b$ estimated using the diffraction maximum that occurs for m = 3 with $d$ = 5 mm and D = 1 m is ______. [2025]

Smart Achievers · Accepted Answer

(94.50)

Given, $\frac{b d}{D} = m λ$ $\Rightarrow b = \frac{D}{d} m λ$

$b = \frac{m λ D}{d} = \frac{3 \times 600 \times 10^{- 3} \times 1}{5 \times 10^{- 3}} μ m = 360 μ m$

$b_{\max} = \frac{3 \times 600 \times 10^{- 3} \times 1.01}{4 \times 10^{- 3}} μ m = 454.5 μ m$

Therefore, absolute error in $b$ ,

$= b_{\max} - b = 454.5 - 360 = 94.50 μ m$

Solution

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Solution

Ans. (94.50) Given, bdD=mλ⇒b=Ddmλ b=mλDd=3×600×10-3×15×10-3 μm=360 μm bmax=3×600×10-3×1.014×10-3 μm=454.5 μm Therefore, absolute error in b, =bmax-b=454.5-360=94.50 μm

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