kadejoset

2022-07-15

A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.

Hassan Watkins

Beginner2022-07-16Added 18 answers

The wavelength of incident light is given as $\lambda =500\times {10}^{-9}\text{}m$

The width of a single slit is given as $a=0.2\times {10}^{-3}\text{}m$

The angular width of central maxima is,

$w=\frac{2\lambda}{a}\phantom{\rule{0ex}{0ex}}=\frac{2\times 500\times {10}^{-9}}{0.2\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=5\times {10}^{-3}m$

Thus, the angular width of the central maxima obtained on the screen is 5 mm.

Given the width of fringe in the young double-slit experiment is $d=0.5\times {10}^{-3}m$

Let there are n fringes made by a double-slit in the central maxima of the diffraction,

Then,

$n=\frac{{B}^{\prime}}{B}$

where ${B}^{\prime}=\frac{2\lambda D}{a}$ is the width of central maxima in the diffraction and $B=\frac{\lambda D}{d}$ is the fringe width by the double slit.

Substituting the known values,

$n=\frac{(\frac{2\lambda D}{a})}{(\frac{\lambda D}{d})}\phantom{\rule{0ex}{0ex}}=\frac{2\lambda D}{a}\times \frac{d}{\lambda D}\phantom{\rule{0ex}{0ex}}=2\frac{d}{a}\phantom{\rule{0ex}{0ex}}=\frac{2\times 0.5\times {10}^{-3}}{0.2\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=5$

Thus, the number of fringes obtained is 5.

The width of a single slit is given as $a=0.2\times {10}^{-3}\text{}m$

The angular width of central maxima is,

$w=\frac{2\lambda}{a}\phantom{\rule{0ex}{0ex}}=\frac{2\times 500\times {10}^{-9}}{0.2\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=5\times {10}^{-3}m$

Thus, the angular width of the central maxima obtained on the screen is 5 mm.

Given the width of fringe in the young double-slit experiment is $d=0.5\times {10}^{-3}m$

Let there are n fringes made by a double-slit in the central maxima of the diffraction,

Then,

$n=\frac{{B}^{\prime}}{B}$

where ${B}^{\prime}=\frac{2\lambda D}{a}$ is the width of central maxima in the diffraction and $B=\frac{\lambda D}{d}$ is the fringe width by the double slit.

Substituting the known values,

$n=\frac{(\frac{2\lambda D}{a})}{(\frac{\lambda D}{d})}\phantom{\rule{0ex}{0ex}}=\frac{2\lambda D}{a}\times \frac{d}{\lambda D}\phantom{\rule{0ex}{0ex}}=2\frac{d}{a}\phantom{\rule{0ex}{0ex}}=\frac{2\times 0.5\times {10}^{-3}}{0.2\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=5$

Thus, the number of fringes obtained is 5.

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