Question

In a double-slit experiment, the slit separation is 2.0 mm, and two wavelengths, 750 nm and 900 nm, illuminate the slits. A screen is placed 2.0 m from the slits. At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other?

Answer 1

Answer:

4.5 x 10⁻6m

Explanation:

From the formula given,

ʎ/d = x/L

The wavelength of 750 nm will be: 750 x 10⁻⁹ / 2.0 x 10⁻³ 3 =  x/ 2.0

Now

50 x 10⁻⁹ / 2.0 x 10⁻³ 3 =  x 2.0  = x = 7.5 x 10⁻⁴m

The wavelength of 900 nm  will be:  900 x 10 ⁻⁹ / 2.0 x 10⁻³ = x /2.0

00 x 10 ⁻⁹ / 2.0 x 10⁻³ = x /2.0 = x = 9.0 x 10⁻⁴

Then

9.0  x 10⁻⁴ / 7.5 x 10⁻⁴ = 1.2

to get to the first integer, 1.3 must be 1.2  is the multiplied  5 times t (which will be a fringe (bright)) which is 6.

Therefore, what this mean is there will be 6 fringes of wavelength 750 nm and 5 fringes of wavelength 900 nm .

From the central brightest fringe.  the mixing fringe will be  4.5 x 10⁻6m  or 4500 nm .

Answer 2

The distance from the central maximum on the screen is 4.5 x 10⁻6m

Calculation of the distance:

Since the slit separation is 2.0 mm, and two wavelengths, 750 nm and 900 nm, illuminate the slits. A screen is placed 2.0 m from the slits

Now

ʎ/d = x/L

So,

750 x 10⁻⁹ / 2.0 x 10⁻³ 3 =  x/ 2.0

Now

50 x 10⁻⁹ / 2.0 x 10⁻³ 3 =  x 2.0  = x = 7.5 x 10⁻⁴m

Now

The wavelength of 900 nm  should be

=  900 x 10 ⁻⁹ / 2.0 x 10⁻³ = x /2.0

= 00 x 10 ⁻⁹ / 2.0 x 10⁻³ = x /2.0 = x = 9.0 x 10⁻⁴

= 9.0  x 10⁻⁴ / 7.5 x 10⁻⁴ = 1.2

=  4.5 x 10⁻6m  or 4500 nm .

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