Question

28. On Introducing a thin film in the path of one of the
two interfering beams in Young's double slit
experiment the central bright fringe shifts to the first
dark fringe. The refracting index of material of film
is 1.6 and wavelength of light used is 6000 A. Then
the thickness of film is
(1) 3 x 10-7m (2) 4 x 10-7m
31 5 x 10-7m
2.5 x 10-7m​

Answer 1

Answer:

thickness of the film is given as

$t = 5 \times 10^{-7} m$

Explanation:

As we know that the shift of the central bright fringe is equal to the half width of one fringe as we know that it shifts to first dark fringe

So we have

$shift = \frac{\lambda D}{2d}$

now we know that the formula for shift of the central bright fringe is given as

$\frac{yd}{D} - (\mu - 1)t = 0$

$y = \frac{(\mu - 1)tD}{d}$

so we have

$\frac{(\mu - 1)tD}{d} = \frac{\lambda D}{2d}$

so we have

$t = \frac{\lambda}{2(\mu - 1)}$

$t = \frac{6000 \times 10^{-10}}{2(1.6 - 1)}$

$t = 5 \times 10^{-7} m$

#Learn

Topic : Young's double slit experiment

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