# Light with a wavelength of 451 nm falls onto a slit of 0.10 mm wide, casting a diffraction pattern onto a screen 3.50 m away. How wide is the central maximum?

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justaguide | Certified Educator

When monochromatic, coherent light passes through a single slit a series of bright and dark fringes are created on a screen on the other side.

The position of the minima are given by the relation `m*lambda = a*sin theta_m` where `lambda` is the wavelength of light, and a is the width of the slit.

Here, the screen is placed at a distance of 3.5 m. This allows `sin theta_m` to be approximated by `y_m/L` . The width of the central maximum is given by the distance between `y_1` and `y_(-1)` . This is equal to `2(L*lambda)/a`

Substituting L = 3.5 m, `lambda` = 451*10^-9 m and a = 0.1*10^-3, the width of the central maximum is 2*3.5*451*10^-9/(0.1*10^-3) = 7*10^-5 m

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