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Wave Optics

Section: Physics

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Question : 2 of 18

Marks: +1, -0

A single slit diffraction experiment is performed to determine the slit width using the equation,

\frac{b d}{D} = m \lambda

b

is the slit width,

D

the shortest distance between the slit and the screen,

d

the distance between the

m^{\text{th}}

diffraction maximum and the central maximum, and

\lambda

is the wavelength.

D

d

are measured with scales of least count of 1 cm and 1 mm , respectively. The values of

\lambda

m

are known precisely to be 600 nm and 3 , respectively. The absolute error (in

\mu \mathrm{m}

) in the value of

b

estimated using the diffraction maximum that occurs for

m=3

d=5\,\mathrm{mm}

D=1\,\mathrm{m}

[JEE Adv 2025 P1]

Your Answer:

Solution:

b = \frac{m \lambda D}{d}

b = \frac{3 \times 600 \times 1}{5} \times 10^{-6}

Case 1 with single dash and case 2 with double dash

\begin{array}{c} b^{'} = \frac{3 \times 600 \times 1.01}{4} \times 10^{-6} \; \frac{N^{+}}{D^{-}} \\ b^{''} = \frac{3 \times 600 \times 0.99}{6} \times 10^{-6} \; \frac{N^{-}}{D^{+}} \end{array}

\begin{array}{c} \frac{b^{'}}{b} = \frac{5.05}{4} \\ \frac{b^{''}}{b} = \frac{4.95}{6} \end{array}

\begin{array}{c} (\Delta b)_1 = \left( \frac{5.05}{4} - 1 \right) b \\ = \frac{1.05}{4} b \\ (\Delta b)_2 = \left( \frac{4.95}{6} - 1 \right) b = \frac{1.05}{6} b \end{array}

\Delta b_1 = \frac{1.05}{4} \times \frac{3 \times 600}{5} \times 10^{-6}

= \frac{90 \times 1.05 \times 10^{-6}}{9450}

\Delta b_1 = 94.50 \, \mu \mathrm{m}

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