Answer:
[tex]80.6\mu m[/tex]
Explanation:
When light passes through a narrow slit, it produces a diffraction pattern on a distant screen, consisting of several bright fringes (constructive interference) alternated with dark fringes (destructive interference).
The formula to calculate the position of the m-th maximum in the diffraction pattern produced in the screen is:
[tex]y=\frac{m\lambda D}{d}[/tex]
where
y is the distance of the m-th maximum from the central maximum (m = 0)
[tex]\lambda[/tex] is the wavelength of light used
D is the distance of the screen from the slit
d is the width of the slit
In this problem, we have:
[tex]\lambda=578.0 nm = 578\cdot 10^{-9} m[/tex] is the wavelength
D = 62.5 cm = 0.625 m is the distance of the screen
We know that the distance between the third order maximum (m=3) and the central maximum is 1.35 cm (0.0135 m), which means that
[tex]y_3 = 0.0135 m[/tex]
For
m = 3
Therefore, rearranging the equation for d, we find the width of the slit:
[tex]d=\frac{m\lambda D}{y_3}=\frac{(3)(578\cdot 10^{-9})(0.625)}{0.0135}=80.3\cdot 10^{-6} m=80.6\mu m[/tex]
The width a of the slit will be "80.6 μm". To understand the calculation, check below.
According to the question,
Light's wavelength, λ = 578.0 nm
Screen's distance, D = 62.5 or,
= 0.625 m
Third order maximum, m = 3
Central maximum = 1.35 cm or,
= 0.0135 m
We know the relation,
→ y = [tex]\frac{m \lambda D}{d}[/tex]
or,
→ Width, d = [tex]\frac{m \lambda D}{y_3}[/tex]
By substituting the values,
= [tex]\frac{3\times 578.10^{-9}\times 0.625}{0.0135}[/tex]
= 80.3 × 10⁻⁶ m or,
= 80.6 μm
Thus the above answer is correct.
Find out more information about wavelength here:
https://brainly.com/question/21630569
