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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics3 Marks
Question
In a single slit diffraction pattern, the distance between first minima on the right and first minima on the left of central maximum is \(4 mm\). The screen on which the pattern is displaced, is \(2 m\) from the slit and wavelength of light used is 6000Å. Calculate width of the slit and width of the central maximum.
✓

Answer

Given:- $\lambda=6000 \AA=6 \times 10^{-7} m$,
$D=2 m$
$X_1+X_2=4 mm=4 \times 10^{-3} m$
To find:- Width of slit (a)
Width of central maximum (W)
Formula:-
i. $X_1+X_2=\frac{2 \lambda D}{a}$
ii. $W=X_1+X_2$
Calculation: From formula (i),
$4 \times 10^{-3}=\frac{2 \times 6 \times 10^{-7} \times 2}{a}$
$\therefore a=6 \times 10^4 m$
The width of the slit is $6 \times 10^{-4} m$.
From formula (ii),
Width of central maximum $=4 \times 10^{-3} m$
The width of the central maximum is $4 \times 10^{-3} m$.

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