Question

In
a double slit experiment light of wavelength 515.0 nm is used, the
separation of the slits is 0.100 mm and the viewing screen is 17.0
cm from the slits. On this screen, what is the separation between
the fifth maximum and seventh minimum from the central
maximum?

Answer #1

wavelength = 515.0 nm =
515 x10 ^{-9} m

the separation of the slits d = 0.100 mm = 0.1x10 ^{-3}
m

Distance of the scrren from the slits D = 17.0 cm = 0.17 m

the separation between the fifth maximum from the central maximum y = ?

Condition of fifth maximum is d sin = 5

For small angles sin ~ tan = y/D

d(y/D) = 5

y = 5D / d

= 5(515 x10 ^{-9} )(0.17) /(0.1x10 ^{-3} )

= 4.4 x10 ^{-3} m = 4.4 mm

Condition of seventh maximum is d sin = 7

For small angles sin ~ tan = y '/D

d(y'/D) = 7

y ' = 7D / d

= 7(515 x10 ^{-9} )(0.17) /(0.1x10 ^{-3} )

= 6.1 x10 ^{-3} m = 6.1 mm

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