Respuesta :
Answer:
2.2 µm
Explanation:
For constructive interference, the expression is:
[tex]d\times sin\theta=m\times \lambda[/tex]
Where, m = 1, 2, .....
d is the distance between the slits.
Given wavelength = 597 nm
Angle, [tex]\theta[/tex] = 15.8°
First bright fringe means , m = 1
So,
[tex]d\times sin\ 15.8^0=1\times \597\ nm[/tex]
[tex]d\times 0.2723=1\times \597\ nm[/tex]
[tex]d=2192.43481\ nm[/tex]
Also,
1 nm = 10⁻⁹ m
1 µm = 10⁻⁶ m
So,
1 nm = 10⁻³ nm
Thus,
Distance between slits ≅ 2.2 µm
Answer:
The separation between the slit is [tex]2.19\mu m[/tex]
Solution:
As per the question:
Wavelength of light, [tex]\lambda = 597 nm = 597\times 10^{-9} m[/tex]
[tex]\theta = 15.8^{\circ}[/tex]
Now, by Young's double slit experiment:
[tex]xsin\theta = n\lambda[/tex]
here,
n = 1
x = slit width
Therefore,
[tex]x = \frac{597\times 10^{-9}}{sin15.8^{\circ}} = 2.19\times 10^{- 6} m[/tex]
[tex]x = 2.19\mu m[/tex]
