Answer to Question #146038 in Optics for Water

Question

A convex lens of focal length 20 cm is placed after a slit of width 0.6 mm. If a plane wave of wavelength 600 nm falls normally on a slit, calculate the separation between the second minima on either side of the central maximum. Calculate the ratio of the intensity of the principal maximum to the first maximum on either side of the principal maximum.

Answer 1

The condition of diffraction minima is given by the following equation.

$bsinθ = mλ$

b – is the split width

m – is the order of diffraction

λ is the wavelength

θ is the diffraction angle

$θ = sin^{-1}(\frac{mλ}{b})$

m = 2

b = 0.6 mm

λ = 600 nm

$θ = sin^{-1}(\frac{2 \times 600 \times 10^{-7}}{0.6 \times 10^{-1}}) = 0.002 \;rad$

The angular position of second order minima on either side of the central maxima is twice that of angle θ.

θ’ = 2θ

$θ’ = 2 \times 0.002 = 0.004\;rad$

The separation between the second minima from central maxima is

d = θ’f

$d = 0.004 \times 20 = 0.08 \;cm$

The intensity distribution is given by the following equation.

$I = I_0(\frac{sin^2β}{β^2})$

I₀ is the intensity of principle maximum.

For the first maximum, the value of β is equal to 1.43π.

β = 1.43π

$I_1 = I_0(\frac{sin^2(1.43π)}{(1.43π)^2}) = 0.0472I_0$

Calculate the ratio of intensities of the principle maximum to first order maximum as follows:

$\frac{I_0}{I_1} = \frac{I_0}{0.0472I_0} = 21$

The ratio of intensities of the principle maximum to first order maximum is 21.