What is the width of a slit for which the first minimum is at 45° when the slit is illuminated by a helium-neon laser (λ = 633 nm) ?

Hint: The small-angle approximation is not valid at 45°.

d = 895 nm

Explanation:

Given:

- Wavelength of light λ = 633 nm

- Angle Q is between central order = 45 degrees

Find:

What is the width of a slit?

Solution:

- The relationship between and wavelength and the order of fringes with respect to slit size d is given by Young's experiment as follows:

sin (Q) = m*λ / d

- Use the above result and compute for the grating d for m = 1,

d = λ / sin (Q)

d = 633*10^-9 / sin (45)

d = 0.89 um

- The width of slit is d = 895 nm