Question

Monochromatic electromagnetic radiation with wavelength λ from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit.

Part A

If the width of the central maximum is 6.00 mm, what is the slit width a if the wavelength is 500 nm (visible light)?

a =

m

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Part B

If the width of the central maximum is 6.00 mm, what is the slit width a if the wavelength is 50.0μm (infrared radiation)?

a =

cm

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Part C

If the width of the central maximum is 6.00 mm, what is the slit width a if the wavelength is 0.500 nm (x rays)?

a =

m

Answer 1

Using the equation

d*sin A = m*lambda

since angle is very small, so

sin A = y/L

d*y/L = m*lambda

y = m*lambda*L/d

Part A:

given values are:

m = 1

lambda = 500 nm

d = 6.00 mm/2 = 3*10^-3 m

L = 2.5 m

So,

y = 2.5*500*10^-9/(3*10^-3) = 4.16*10^-4 m

Part B:

given values are:

m = 1

lambda = 50 um = 50*10^-6 m

d = 6.00 mm/2 = 3*10^-3 m

L = 2.5 m

So,

y = 2.5*50*10^-6/(3*10^-3) = 4.16*10^-2 m = 4.16 cm

Part C:

given values are:

m = 1

lambda = 0.5 nm

d = 6.00 mm/2 = 3*10^-3 m

L = 2.5 m

So,

y = 2.5*0.5*10^-9/(3*10^-3) = 4.16*10^-7 m