For normal incidence in the far field, d is slit spacing, lambda wavelength, n order and theta the angle from the grating normal. Which equation gives principal reinforcement directions?

• d cos theta equals n lambda
• d sin theta equals n lambda
• d tan theta equals n lambda

A grating has 500 lines per millimetre. What is its centre-to-centre slit spacing d?

• 2 × 10^−6 metres
• 5 × 10^−1 metres
• 6 × 10^−7 metres

For normal incidence, an order requiring sin(theta) greater than 1 can occur at an angle beyond 90 degrees, with theta measured from the grating normal.

For normal incidence, d = 2 × 10^−6 m and lambda = 600 nm. What is the largest positive order permitted by the sine bound? Include a grazing direction if the ratio is exactly an integer.

For normal incidence in the far field, d = 3.0 × 10^−6 m and lambda = 450 nm. For order n = 2, find sin(theta), with theta measured from the grating normal.

With spacing and wavelength fixed, why does increasing the number of illuminated grating slits narrow the principal maxima?

• More slits reduce the wavelength passing through the grating
• More slits increase the spacing between neighbouring slits
• More slits increase cancellation away from principal maxima

Ava calculates 1/250 = 0.004 for a grating with 250 lines per millimetre and writes d = 0.004 m. Which step needs correction?

• Replace reciprocal density with diffraction order
• Replace reciprocal density with wavelength squared
• Convert the reciprocal spacing from mm to metres

For normal incidence in the far field, order n = 2 has sin(theta) = 0.625, with theta measured from the grating normal. The spacing is d = 1.6 × 10^−6 m. Find lambda in metres.