Physics, asked by neelamseth1974ns, 12 hours ago

In a biprism experiment, the separation of the slits is halved and the distance between the slits and the screen is doubled. How is the fringe width affected?​

Answers

Answered by chakrichakraverthi
0

Answer:

Separation between the slit = 1/2

Distance between the slits and screen = 2x

Young's double slit experiment on interference of light has the formula -

X = λD/d

where X is the band width or fringe width, A is the wave length of the monochromatic light, D is the distance between the slits and screen, and d is the distance

between the slits.

Maxima = (n+1) XD/d

Yn = Distance of nth

Maxima = n^D/d

Therefore, D2 = 2D1

= D2/D1 x d1/d2

= 2x2

= 4

Thus, If the distance between the slits is doubled, the fringr width is is quadrupled.

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