Question

In a single slit diffraction with λ=500nm and a lens of diameter 0.1 mm, width of central maxima, obtain on screen at a distance of 1 m will be
A. 5 mm
B. 1 mm
C. 10 mm
D. 2.5 mm

Answer 1

We are given the wavelength of the light and diameter of the lens used in a single slit experiment. We know that the width of the central maxima is the distance between the two minima. Therefore by finding the angle subtended by two minima and substituting it in the equation we will get the required solution.

Formula used:
$\alpha =\dfrac{2\lambda }{w}$
$x=d\alpha$

Complete step-by-step answer:
In the question we are given a single slit diffraction set up.
The wavelength of the light is given as,
$\lambda =500nm$
Wavelength is given to us in nanometer units; let us convert it into meters. Thus we get,
$\lambda =500\times {{10}^{-9}}m$
The diameter of the lens is given as 0.1 mm.
In this case we can consider the diameter of the lens as the width of the slit.
Therefore we can say width,
$w=0.1mm$
Since it is given in millimeters, let us convert this also into meters.
$w=0.1\times {{10}^{-3}}$
We know that the angle subtended by two minima at the slit in a single slit diffraction is given by,
$\alpha =\dfrac{2\lambda }{w}$, where ‘$\lambda$’ is the wavelength and ‘w’ is the width of the slit.
Let us substitute for known values in the above equation, thus we get
$\Rightarrow \alpha =\dfrac{2\times \left( 500\times {{10}^{-9}} \right)}{\left( 0.1\times {{10}^{-3}} \right)}$
By solving this we get,
$\Rightarrow \alpha =0.01m$
In the question we are asked to find the width of central maxima obtained on the screen at a distance of 1 m, i.e.
The distance between the slit and the screen is given as,
$d=1m$
We know that the width of central maxima is the distance between the two minima, which is given as,
$x=d\alpha$
Since we know the values of ‘$d$’ and ‘$\alpha$’ we can substitute that in the above equation. Thus we get,
$\Rightarrow x=1m\times 0.01m$
$\Rightarrow x=0.01m$
Since all the options are given in millimeters, we can convert this into millimeters. Thus we get,
$\Rightarrow x=0.01\times {{10}^{3}}$
$\Rightarrow x=10mm$
Therefore the width of the central maxima is 10 mm.

So, the correct answer is “Option C”.

Note: Diffraction is defined as the tendency of a wave spread out as it propagates when it is emitted from a finite source or is passing through a small aperture. Interference is the result of diffraction when a finite number of waves are emitted from a source.
In a single slit experiment when we pass light through a single slit we observe a single slit diffraction pattern on the screen which is at a distance from the slit.