In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
[tex]y_n= \frac{n \lambda D}{a} [/tex] (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
[tex]\lambda[/tex] is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
The width of the first peak (let's call it [tex]\Delta y[/tex]) corresponds to twice the distance of the first minimum:
[tex]\Delta y = 2 y_1=9.90 cm=0.099 m[/tex]
which means
[tex]y_1 = \frac{\Delta y}{2}= \frac{0.099 m}{2}=0.0495 m [/tex]
The wavelength of the light in the problem is
[tex]\lambda=415 nm=4.15 \cdot 10^{-7} m[/tex]
and the distance of the screen from the slit is
[tex]D=2.53 m[/tex]
So by using these data and re-arranging the equation, we can find the width of the slit:
[tex]a= \frac{1 \lambda D}{y_1} = \frac{(1)(4.15 \cdot 10^{-7} m)(2.53 m)}{0.0495 m}=2.15 \cdot 10^{-5} m = 0.0215 mm [/tex]
