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  4. Light of wavelength 5000 mathringA is incident normally on a slit. First minimum of diffraction pattern is formed at a distance of 5 mm from the central maximum. If slit width is 0.2 mm, then distance between slit and screen will be :

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Q. Light of wavelength $5000 \,\mathring{A}$ is incident normally on a slit. First minimum of diffraction pattern is formed at a distance of $5\, mm$ from the central maximum. If slit width is $0.2 \,mm$, then distance between slit and screen will be :

Wave Optics

Solution:

Diffraction minima on either side of central maxima is given as
$b \sin \theta=n \lambda$
$\Rightarrow b \cdot \frac{x}{D}=n \lambda$
$D =\frac{b x}{n \lambda}=\frac{2 \times 10^{-4} \times 5 \times 10^{-3}}{1 \times 5000 \times 10^{-10}}=2.0\, m$