Question

Light of wavelength 610 nm falls on a slit that is 3.50×10^−3 mm wide. How far the first bright diffraction fringe is from the strong central maximum if the screen is 10.0 m away.
Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

2.6143 m

Explanation:

For constructive interference, the expression is:

$d\times sin\theta=m\times \lambda$

d is the distance between the slits.

The formula can be written as:

$sin\theta=\frac {\lambda}{d}\times m$ ....1

The location of the bright fringe is determined by :

$y=L\times tan\theta$

Where, L is the distance between the slit and the screen.

For small angle , $sin\theta=tan\theta$

So,  

Formula becomes:

$y=L\times sin\theta$

Using 1, we get:

$y=L\times \frac {\lambda}{d}\times m$

For the single slit diffraction, the bright fringes are represented by the half-integers. The first such integer is, m = 1.5

y = ?

Given L = 10.0 m

d = 3.50 × 10⁻³ mm

Also, 1 mm = 10⁻³ m

So, d = 3.50 × 10⁻⁶ m

λ = 610 nm  

Since, 1 nm = 10⁻⁹ m

So,  

λ = 610 × 10⁻⁹ m

Applying the formula as:

$y=10.0\ m\times \frac {610\times 10^{-9}\ m}{3.50\times 10^{-6}\ m}\times 1.5$

⇒ y, location of first bright fringe = 2.6143 m