1. Coherent light of wavelength 500nm is incident on two very narrow slits. The

Question

1. Coherent light of wavelength 500nm is incident on two very narrow slits. The light has the same phase at both slits.

a) What is the angular separation in radians between the central maximum in intensity and an adjacent maximum if the slits are 0.01mm apart?

b) What would be the separation between maxima in intensity on a screen located 1m from the slits?

c) What would the angular separation be if the slits were 1mm apart?

d) What would the separation between maxima be on a screen 25mm from the slits? The answer is relevant to the maximum resolution along your retina that would be useful.

e) How would the location of the maxima change if one slit was covered by a thin film with higher index of refraction that shifted the phase of light leaving the slit by ? Would the separation change?

Explanation / Answer

As we know that in double slit interfrence

a*sin(theta) = m * lambda

So,

a)

theta = m*lambda / a ............(Because theta is very small)

theta = 500 * 10^-9 / 0.01*10^-3

theta = 0.05 degree

b)

y = m * lambda * D / d

y = 500 * 10^-9 * 1 / 0.01 * 10^-3 = 0.05 m

c)

theta = m*lambda / a ............(Because theta is very small)

theta = 500 * 10^-9 / 1*10^-3

theta = 5 * 10^-4 degree

d)

y = m * lambda * D / d

y = 500 * 10^-9 * 25 * 10^-3 / 0.01 * 10^-3 = 1.25 * 10^-3 m

e)

y = m * (lambda/n) * D / d

n = refractive index of film

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