Answer:
0.145 m
Explanation:
Data provided in the question:
wavelength of red light, λred = 656 nm = 656 × 10⁻⁹ m
wavelength of blue light, λblue = 486 nm = 486 × 10⁻⁹ m
line density = 500 per mm
length, L = 1.3 m
d = 1 mm / 500 lines
= 0.002 mm = 0.002 × 10⁻⁶ m
m = 1 [for the first order bright fringe]
Now,
The positions can be determined by using the formula
θ = [tex]\sin^{-1}(\frac{m\lambda}{d})[/tex]
& y = Ltan(θ)
thus,
θred = [tex]\sin^{-1}(\frac{1\times656\times10^{-9}}{0.002\times{10^{-6}}})[/tex]
= 19.15°
yred = Ltan(θred)
= (1.5) × tan(19.15°)
= 0.521 m
similarly,
θblue = [tex]\sin^{-1}(\frac{1\times486\times10^{-9}}{0.002\times{10^{-6}}})[/tex]
= 14.1°
yblue = Ltan(θblue)
= 1.3 × tan(14.1°)
= 0.376 m
Hence,
distance between the first-order red and blue fringes
= 0.521 m - 0.376 m
= 0.145 m