Question 346: Waves Optics

Question

Light of wavelength ',$$\lambda$$' is incident on a slit of width '$$\mathrm{d}$$'. The resulting diffraction pattern is observed on a screen at a distance '$$D$$'. The linear width of the principal maximum is then equal to the width of the slit if $$D$$ equals

Accepted Answer

In diffraction of light by single slit, the width of central maximum is given as
$$\mathrm{W}_{\mathrm{c}}=\frac{2 \lambda \mathrm{D}}{\mathrm{d}}$$
Given: $$\mathrm{W}_{\mathrm{c}}=\mathrm{d}$$
$$\begin{aligned}
	herefore \quad d & =\frac{2 \lambda D}{d} \
\Rightarrow D & =\frac{d^2}{2 \lambda}
\end{aligned}$$

Answer

$$\frac{d}{\lambda}$$

Answer

$$\frac{\mathrm{d}^2}{2 \lambda}$$

Answer

$$\frac{2 \lambda}{\mathrm{d}}$$

Answer

$$\frac{2 \lambda^2}{d}$$

Waves Optics

Light of wavelength ',$\lambda' is incident on a slit of width '\mathrm{d}'. The resulting diffraction pattern is observed on a screen at a distance 'D'. The linear width of the principal maximum is then equal to the width of the slit if D$ equals

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