Question

Monochromatic light of wavelength λ is incident on a pair of slits separated by 2.30 10-4 m and forms an interference pattern on a screen placed 2.10 m from the slits. The first-order bright fringe is at a position ybright = 4.56 mm measured from the center of the central maximum. From this information, we wish to predict where the fringe for n = 50 would be located.

(a) Assuming the fringes are laid out linearly along the screen, find the position of the n = 50 fringe by multiplying the position of the n = 1 fringe by 50.0.

(b) Find the tangent of the angle the first-order bright fringe makes with respect to the line extending from the point midway between the slits to the center of the central maximum.

(c) Using the result of part (b) and dsinθbright = mλ, calculate the wavelength of the light. nm

(d) Compute the angle for the 50th-order bright fringe from dsinθbright = mλ. °

(e) Find the position of the 50th-order bright fringe on the screen from ybright = Ltanθbright.

Answer 1

Step-by-step explanation:

fringe width = λ D / d , λ is wavelength of light , D is distance of screen , d is slit width.

first bright fringe is located at 4.56 mm

so fringe width = 4.56 mm

location of 50 th fringe = 50 x 4.56

= 228 mm .

b ) If angular position of first order fringe be Ф

tanФ = λ / d

= fringe width / D , D is distance of screen

= 4.56 / 2.1 x 1000

4.56 / 2100

c )

d sinФ = λ ( wavelength)

= 2.30 x 10⁻⁴ x 4.56 / 2100 = wavelength [ when angle is small , sinФ = tanФ ]

= 499.4 nm .

d ) d sinθ = 50 x 499.4 x 10⁻⁹

sinθ = 50 x 499.4 x 10⁻⁹ / 2.30 x 10⁻⁴

= .2497 / 2.3

= 6.23 degree .

e ) ybright = Ltanθbright , L is distance of screen

= 2.1 x tan6.23

= .229 m

= 229 mm .