A plane wave of wavelength $590 n m$ is incident on a slit with a width of $a = 0.40 m m .$ A thin converging lens of focal length $+ 70 c m$ is placed between the slit and a viewing screen and focuses the light on the screen.
What is the distance on the screen from the center of the diffraction pattern to the first minimum?

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Solution

Waves leaving the lens at an angle $θ$ to the forward direction interfere to produce an intensity minimum if $a sin θ = m λ,$ where $a$ is the slit width, $λ$ is the wavelength, and $m$ is an integer.
The distance on the screen from the center of the pattem to the minimum is given by $y = D tan θ$ , where $D$ is the distance from the lens to the screen.
For the conditions of this problem,
$sin θ = \frac{m λ}{a} = \frac{(1) (590 \times 10^{- 9} m)}{0.40 \times 10^{- 3} m} = 1.475 \times 10^{- 3}$
This means $θ = 1.475 \times 10^{- 3} r a d$ and
$y = (0.70 m) tan (1.475 \times 10^{- 3} r a d) = 1.0 \times 10^{- 3} m$

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