Question

A diffraction grating has 1 950 rulings/cm. On a screen 1.00 m from the grating, it is found that for a particular order m, the maxima corresponding to two closely spaced wavelengths of sodium (589.0 nm and 589.6 nm) are separated by 0.42 mm. Determine the value of m.

Exact answers with detail please

Answer 1

d = 1/1950 = 5.12 * 10^-6 m
d*sin(θ) = m*λ
θ = sin^-1(m*λ/d)
y = L * tan(θ)
∆y = L * [ tan(sin^-1(m*λ2/d)) - tan(sin^-1(m*λ1/d)) ]

For m = 1,
∆y = 1.0 *  [ tan(sin^-1( (1*589.6 * 10^-9)/(5.12 * 10^-6) ) ) - tan(sin^-1( (1*589 * 10^-9)/(5.12 * 10^-6))) ]
∆y = 0.12 mm

For m = 2,
∆y = 1.0 *  [ tan(sin^-1( (2*589.6 * 10^-9)/(5.12 * 10^-6) ) ) - tan(sin^-1( (2*589 * 10^-9)/(5.12 * 10^-6))) ]
∆y = 0.25 mm

For m = 3,
∆y = 1.0 *  [ tan(sin^-1( (3*589.6 * 10^-9)/(5.12 * 10^-6) ) ) - tan(sin^-1( (3*589 * 10^-9)/(5.12 * 10^-6))) ]
∆y = 0.42 mm

So Value of m = 3