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Troyanec [42]
3 years ago
14

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

om the slit, when light of wavelength 530 nm is used. find the slit width.
Physics
1 answer:
WARRIOR [948]3 years ago
7 0
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
y_n= \frac{n \lambda D}{a} (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
\lambda is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem, 
D=37.0 cm=0.37 m
\lambda=530 nm=5.3 \cdot 10^{-7} m
while the distance between the first and the fifth minima is
y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m (2)

If we use the formula to rewrite y_5, y_1, eq.(2) becomes
\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m
Which we can solve to find a, the width of the slit:
a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}=  1.57 \cdot 10^{-3} m=1.57 mm

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