Question

Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark bands on the screen are measured to be 1.70 cm apart. What is the wavelength of the incident light? (A) 612 nm (B) 457 nm (C) 306 nm (D) 392 nm

Answer 1

Answer:

Wavelength of incident light, $\lambda = 612 nm$

Given:

Distance between slit and screen, x = 5.00 m

slit width, d = 0.180 mm

width of the fringe, $\beta = 1.70 cm = 0.017 m$

Solution:

To calculate the wavelength of the incident light, $\lambda$:

$\beta = \frac{x\lambda }{d}$

$\lambda = \frac{\beta d}{x}$

$\lambda = \frac{0.017\times 0.180\times 10^{- 3}}{5} = 6.12\times 10^{- 7}m = 612 nm$

$\lambda = 612 nm$