Question

A beam of light of wavelength 420 nm, is used to obtain interference fringes in

a Young’s double-slit experiment. Find the distance of the third dark fringe on the

screen from the central maxima. Take the separation between the slits as 4.2 mm

and the distance between the screen and plane of the slits as 1.4 m.

Section ​

Answer 1

Answer:

Distance of the third dark fringe = 350 × 10⁻⁶ m.

Explanation:

Given:

• Wavelength of light = 420 nm = 420 × 10⁻⁹m
• Separation between slits = 4.2 mm = 4.2 × 10⁻³m
• Distance between screen and slit = 1.4 m

To Find:

• Distance of the third dark fringe on the screen

Solution:

The position of the nth dark fringe from the central maxima is given by,

$\boxed{\tt y_n=(2n-1)\times \dfrac{D\lambda}{2d}}$

where D = distance of the screen from the slit

d = separation between the slit

λ = wavelength of the light

n = 1, 2, 3..

Substituting the given data we get,

$\tt y_3=(2\times 3-1)\times \dfrac{1.4\times 420\times 10^{-9}}{2\times 4.2\times 10^{-3}}$

$\tt y_3=(6-1)\times \dfrac{1.4\times 420\times 10^{-9}\times 10^3}{2\times 4.2}$

$\tt y_3=5\times \dfrac{1.4\times 420\times 10^{-6}}{8.4}$

$\tt y_3= \dfrac{2940\times 10^{-6}}{8.4}$

$\tt y_3= 350\times 10^{-6}\:m$

Hence the distance of the third dark fringe is 350 × 10⁻⁶ m.

Answer 2

$\longrightarrow$ REQUIRED ANSWER :-

$y3 = (2 \times 3 - 1) \times \frac{1.4 \times 420 \times 10 {}^{3} }{2 \times 4.2 \times 10 {}^{ - 2} }$

$y3 = (6 - 1) \times \frac{1.4 \times 420 \times 10 {}^{ - 9 \: } } {2 \times 4.2}$

$y3 = 5 \times \frac{1.4 \times 420 \times 10 {}^{2} }{8.4}$

$y3 = \frac{2940 \times 10 {}^{ - 2} }{8.4}$

$y3 = 350 \times 10 {}^{ - 2} m$

$\huge\underbrace\mathfrak\pink{ᴀɴsᴡᴇʀ}$

$305 \times 10 {}^{ - 2}$