Question

Light of wavelength 440 nm in air shines on two slits 5.50×10−2 mm apart. The slits are immersed in water (n = 1.33), as is a viewing screen 45.0 cm away.

How far apart are the fringes on the screen?

Answer 1

In double slit experiment,

d*sin theta = m*lambda

also, tan = y/D

Since theta is very small So,

sin theta = tan theta
m*lambda/d = y/D

therefore position of maxima is given by,

y = m*lambda*D/d

So, linear distance on the screen between adjacent fringes will be = dy = y(m+1) - y(m)

here, m = order of fringes

lambda0 = wavelength in air = 440 nm = 440*10^-9 m

lambda = wavelength in water = lambda0/n

n = refractive index of water = 1.33

lambda = 440*10^-9/1.33

D = slit-screen separation = 45.0 cm = 0.45 m

d = slit separation = 5.50*10^-2 mm = 5.50*10^-5 m

So,

dy = (m+1)*lambda*D/d - m*lambda*D/d

dy = lambda*D/d

dy = (440*10^-9)*0.45/(1.33*5.50*10^-5)

dy = 0.00271 m

dy = 2.71*10^-3 m

dy = 2.71 mm = distance between fringes

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