Light of wavelength 500 nm is used to form interference pattern in Young's double slit experiment. A uniform glass plate of refractive index 1.5 and thickness 0.1 mm is introduced in the path of one of the interfering beams. The number of fringes which will shift the cross wire due to this is

Question

Light of wavelength 500 nm is used to form interference pattern in Young's double slit experiment. A uniform glass plate of refractive index 1.5 and thickness 0.1 mm is introduced in the path of one of the interfering beams. The number of fringes which will shift the cross wire due to this is (A) 100 (B) 200 (C) 300 (D) 400

Tardigrade Admin · Accepted Answer

The number of fringes shifting is decided by the extra path difference produced by introducing the glass plate. The extra path difference is

(μ - 1) t = nλ

or

(1.5 - 1) \times 0.1 \times 1 0^{- 3} = n \times 500 \times 1 0^{- 9}

= n = 100

Q. Light of wavelength 500nm is used to form interference pattern in Young's double slit experiment. A uniform glass plate of refractive index 1.5 and thickness 0.1mm is introduced in the path of one of the interfering beams. The number of fringes which will shift the cross wire due to this is

100

200

300

400

The number of fringes shifting is decided by the extra path difference produced by introducing the glass plate. The extra path difference is (μ−1)t=nλ or (1.5−1)×0.1×10−3=n×500×10−9 =n=100

The number of fringes shifting is decided by the extra path difference produced by introducing the glass plate. The extra path difference is $(μ - 1) t = nλ$ or $(1.5 - 1) \times 0.1 \times 1 0^{- 3} = n \times 500 \times 1 0^{- 9}$
$= n = 100$