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 awa…

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asked Jul 6, 2020 115k views

1 Answer

Dimitri T · Answer 1 · 2020-07-10T21:06:39+0000

Answer:

2.6143 m

Step-by-step explanation:

For constructive interference, the expression is:

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

d is the distance between the slits.

The formula can be written as:

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

The location of the bright fringe is determined by :

$y=L* tan\theta$

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

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

So,

Formula becomes:

$y=L* sin\theta$

Using 1, we get:

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

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:

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

⇒ y, location of first bright fringe = 2.6143 m