If a transparent medium of refractive index μ=1.5 and thickness t=2.5 × 10-5 m is inserted in front of one of the slits of Young's Double Slit experiment, how much will be the shift in the interference pattern? The distance between the slits is 0.5 mm and that between slits and screen is 100 cm

Question

If a transparent medium of refractive index μ=1.5 and thickness t=2.5 × 10-5 m is inserted in front of one of the slits of Young's Double Slit experiment, how much will be the shift in the interference pattern? The distance between the slits is 0.5 mm and that between slits and screen is 100 cm (A) 5 cm (B) 2.5 cm (C) 0.25 cm (D) 0.1 cm

Tardigrade Admin · Accepted Answer

Shift in the fringe pattern x=((μ-1) t . D/d) =((1.5-1) × 2.5 × 10-5 × 100 × 10-2/0.5 × 10-3)=2.5 cm

Q. If a transparent medium of refractive index μ=1.5 and thickness t=2.5×10−5m is inserted in front of one of the slits of Young's Double Slit experiment, how much will be the shift in the interference pattern? The distance between the slits is 0.5mm and that between slits and screen is 100cm

5 cm

2.5 cm

0.25 cm

0.1 cm

Shift in the fringe pattern x=d(μ−1)t.D​ =0.5×10−3(1.5−1)×2.5×10−5×100×10−2​=2.5cm

Shift in the fringe pattern $x = \frac{( μ - 1 ) t . D}{d}$
$= \frac{( 1.5 - 1 ) \times 2.5 \times 1 0 ^{- 5} \times 100 \times 1 0 ^{- 2}}{0.5 \times 1 0 ^{- 3}} = 2.5 c m$