The 3rd bright fringe of a double slit interference pattern is 30.0 cm above the central bright fringe. If the angle from the horizontal to this 3rd bright fringe is 12.0 degrees, what is the distance (in meters) between the double slits and the viewing screen

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Answer:

Explanation:

12 degree = (π / 180) x 12 radian

= .2093 radian

position of third bright fringe = 3λ D/ d where λ is wave length of light , D is screen distance and d is slit separation

given

3λ D/ d = 30 x 10⁻²

angular separation

3λ / d = .2093

from the two equation

.2093 D= 30 x 10⁻²

D = 30 x 10⁻² / .2093

= 1.43 m

The distance between the double slits and the viewing screen will be "1.43 m". To understand the calculation, check below.

Wavelength and Distance

According to the question,

Angle = 12° or,

          = ([tex]\frac{\pi}{180}[/tex]) × 12

          = 0.2093 radian

Double slit interference pattern, d = 30.0 cm

We know the relation,

→ 3λ [tex]\frac{D}{d}[/tex] = 30 × 10⁻² ...(equation 1)

or, The angular separation be:

→      [tex]\frac{3 \lambda}{d}[/tex] = 0.2093 ...(equation 2)

From equation "1" and "2", we get

 0.2093 D = 30 × 10⁻²

               D = [tex]\frac{30\times 10^{-2}}{0.2093}[/tex]

                   = 1.43 m

Thus the above approach is correct.

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