Answer:
The width of the slit is [tex]d = 5.68 *10^{-5} \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 610 \ nm = 610 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 1.23 ^o[/tex]
Generally the angle between the first minimum on one side and that the central maximum is evaluated as
[tex]\theta _1 = \frac{\theta}{2}[/tex]
=> [tex]\theta _1 = \frac{1.23}{2}[/tex]
=> [tex]\theta _1 = 0.615 ^o[/tex]
Generally the condition for minimum diffraction is mathematically represented as
[tex]d sin \theta_1 = n\lambda[/tex]
For first minimum n = 1
=> [tex]d = \frac{n \lambda }{ sin (\theta_1)}[/tex]
=> [tex]d = \frac{1 * 610 *10^{-9}}{ sin (0.615)}[/tex]
=> [tex]d = 5.68 *10^{-5} \ m[/tex]
