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# Two slits are made one millimetre apart with screen placed one metre away and blue-green light of wavelength $500nm$ is used. What should the width of each slit be to obtain $10$ maxima of double slit pat-tern within the central maximum of single slit pattern?$\left( A \right)0.1mm$$\left( B \right)0.05mm$$\left( C \right)0.2mm$$\left( D \right)0.4mm$

Last updated date: 20th Jun 2024
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Hint: We know that the fringe width is the distance between the two dark fringes or bright fringes or two dark fringes successfully.
It is independent of order of all the fringes. Double slit experiment shows both energy and matter show both wave and particle characteristics.
Apply this logic to find the width with central maximum.

Formula used:
$\beta = \dfrac{{\lambda D}}{d}$
Here $d,D,\lambda$ is the distance between the slits, distance between the slit and screen and wavelength.

Two coherent sources of lights placed at a small distance apart are used in Young’s double slit experiment. Usually only magnitudes greater than wavelength of light is used.
Young’s double slit experiment helps in understanding the wave theory of light. A commonly used coherent source in the modern-day experiments is Laser.
Young’s double slit experiment firmly establishes that light behaves as a particle and wave.
The distance between the two bright fringes or two dark fringes successfully is said to fringe width. It is independent of order of all the fringes.
The brightest central fringe from a diffraction pattern on the screen is considered to be the central maxima. It has the maximum intensity than all other diffraction fringes.
Fringe separation is given by $\beta = \dfrac{{\lambda D}}{d}$
Here d is ${10^{ - 3}}m$
And $D = 1m$
Then the width of central maxima in a single slit experiment=$\dfrac{{2\lambda D}}{a}$
Fringe width in double slit experiment $\beta = \dfrac{{\lambda D}}{d}$
Hence, we have to obtain for 10 maxima of double slit
$\Rightarrow 10\beta = \dfrac{{2\lambda D}}{a}$
On putting the values and we get
$\Rightarrow 10\dfrac{{\lambda D}}{d} = \dfrac{{2\lambda D}}{a}$
Then, we can write it as
$\Rightarrow a = \dfrac{d}{5}mm = \dfrac{1}{5}mm = 0.2mm$

Hence Option C is the correct option.

Note: Later they conducted Young’s double slit experiment using electrons and the pattern generated a similar result as light. It behaves both as a particle and wave. Young’s double slit experiment shows both these characteristics prominently.