Answer:
The width of the slit is 0.167 mm
Explanation:
Wavelength of light, [tex]\lambda=608\ nm=608\times 10^{-9}\ m[/tex]
Distance from screen to slit, D = 88.5 cm = 0.885 m
The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m
We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :
[tex]y=\dfrac{mD\lambda}{a}[/tex]
where
a = width of the slit
[tex]a=\dfrac{mD\lambda}{y}[/tex]
[tex]a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}[/tex]
a = 0.000167 m
[tex]a=1.67\times 10^{-4}\ m[/tex]
a = 0.167 mm
So, the width of the slit is 0.167 mm. Hence, this is the required solution.
