In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away.
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.

Solution:

Distance between the slits, $d=0.28 \mathrm{~mm}=0.28 \times 10^{-3} \mathrm{~m}$

Distance between the slits and the screen, = 1.4 m

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

= 1.2 cm = 1.2 × 10−2 m

In case of a constructive interference, we have the relation for the distance between the two fringes as:

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

Where,

n = Order of fringes = 4

λ = Wavelength of light used

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

$=\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}$

$=6 \times 10^{-7}$

$=600 \mathrm{~nm}$

Hence, the wavelength of the light is 600 nm.

Administrator

Leave a comment

Please enter comment.
Please enter your name.