Answer:
(a) 1.029 m
(b) 1.78 mm
Solution:
As per the question:
Wavelength of light, [tex]\lambda = 588\ nm[/tex]
Slit width, w = 0.68 mm
Now,
The distance first minimum in the diffraction pattern, [tex]y_{1} = 0.89\ mm[/tex]
Now,
(a) For first minima:
[tex]wsin\theta = \lambda[/tex]
[tex]w(\frac{y}{D}) = \lambda[/tex]
where
D = Distance from the screen
[tex]D = (\frac{0.68\times 10^{- 3}\times 0.89\times 10^{- 3}}{588\times 10^{- 9}})[/tex]
D = 1.029 m
(b) The width of the central maxima is given by:
2y = [tex]2\times 0.89 = 1.78\ mm[/tex]
