In a double-slit experiment, the third-order maximum (n=3) for light of waveleng

Question

In a double-slit experiment, the third-order maximum (n=3) for light of wavelength 500 nm is located 12 mm (center-to-center) from the central bright spot on a screen L=1.6 m away from the slits.

(A)  What is the separation d between the slits?

(B)  Is the small angle approximation applicable in this problem?

(C)  Light of wavelength 650 nm is then projected through the same slits. Make a table showing how

far the interference minima (dark lines) are located from the center of the interference pattern for n=0,1,2.

Explanation / Answer

a)

d (y/L) = n lamda

d = n lamda L/y = 3*500e-9*1.6/12e-3 = 0.0002 m

b)

yes

because we have: d << L

c)

d (y/L) = ((2n+1)/2) lamda

y = ((2n+1)/2) lamda L/d

==> y = ((2n+1)/2) * 650e-9 * 1.6/0.0002 = 0.0026 * (2 n + 1)

n=0 ==> y = 0.0026 * (2 n +1) = 0.0026 * (2 * 0 + 1) = 0.0026 m

n=0 ==> y = 0.0026 * (2 n +1) = 0.0026 * (2 * 1 + 1) = 0.0078 m

n=0 ==> y = 0.0026 * (2 n +1) = 0.0026 * (2 * 3 + 1) = 0.0182 m

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