Question

If a transparent medium of refractive index mu is equal to 1.5 and thickness t is equal to 2.5 into 10 ki power minus 5 metre is inserted in front of a slit of young double slit experiment how much will be the shift in the interference pattern the distance between the slit is 2.5 mm and that between the slit and the screen is hundred centimetre ​

Answer 1

The shift in the interference pattern will be 2.5 cm

Explanation:

Shift in the fringe pattern x = (μ − 1)  t.D / d

(μ − 1)  t. = x.d / D

x = (μ − 1)  t.d /D

x = (1.5 − 1) × 2.5 × 10^−5 × 100 × 10^−2 / 0.5×10^−3

x = 2.5 cm

Hence the shift in the interference pattern will be 2.5 cm

Also learn more

In Young’s double slit experiment the distance between the slits and the screen is doubled. The separation between the slits is reduced to half. As a result the fringe width(a) is doubled(b) is halved(c) becomes four times(d) remains unchanged?

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