Question

In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Answer 1

Answer:-

★ We know that,

→ Distance between the slits and the screen, D = 1.4 m

→ And, the distance between the slits, d = 0.28 mm = 0.28 x 10-³m

★Now,

→ Distance between the central fringe and the fourth (n = 4) fringe,

u = 1.2cm = 1.2 × 10-² m

★ Remember:-

→ For a constructive interference, following is the relation for distance between the two fringes

$u = n\lambda \frac{D}{d}$

Where, n = order of fringes

→ 4 λ = Wavelength of light used.

$u = n\lambda \frac{D}{d}$

$\frac{1.2 \times {10}^{ - 2} \times 0.28 \times {10}^{ - 3} }{3 \times 1.4}$

= $60\times{10}^{-7} = 600nm$

Therefore, 600nm is the wavelength of the light.