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levacccp [35]
3 years ago
11

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 s

trong central maximum if the screen is 10.0 m away.
Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
romanna [79]3 years ago
3 0

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

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