In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and r…

Question

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n (=1.5 ), the central fringe shifts by 0.2 cm . The value of t is \_ \_ \_ \_ cm.

Accepted Answer

Correct answer: C. (c) Given: d=0.1 cm ,D=50 cm ,n=1.5, Shift =0.2 cm Optical path difference by sheet: = (n-1 )t Shift of central fringe: Shift = (n-1 )t Dd 0.2= (1.5-1 ) t 500.1= 0.5 50 t0.1=250t t= 0.2250=8 10^-4 cm

$In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness and refractive index , the central fringe shifts by 0.2 cm . The value of is cm.$

Solution

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