In an interference experiment with a biprism, the distance of the slits from the screenn is increased by 25% annd the separation between the slits is

Question

In an interference experiment with a biprism, the distance of the slits from the screenn is increased by 25% annd the separation between the slits is

In an interference experiment with a biprism, the distance of the slits from the screenn is increased by 25% annd the separation between the slits is halved. If X represents the original fringe width, the new fringe width will be
A. `2.5X`
B. `1.5X`
C. `2X`
D. `4X`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
`D_(2)=1.25 D_(1) D_(2)=(d_(2))/(2), beta_(2)= b ? beta_(1)=x`
`b=(D)/(d)`
`:. (beta_(2))/(beta_(1))=(D_(2))/(D_(1))xx(d_(1))/(d_(2))`
`beta_(2)=(1.25 D_(1))/(D_(1))xx (d_(2) xx 2)/( d_(1)) xx beta_(1)`
`beta_(2)=2.5 beta_(1) = 2.5 X`

4 views Answered 2023-02-05 15:29