Question

two slits are 1mm apart and screen is placed at some distence .when the slits are illuminated with light of wavelength 500 nm,fringe separation obtained on the screen is 0.5mm.what is the distence between the screen and the slits?​

Answer 1

◽The distance between the screen and the slits.⬇

$given.. \\ λ = 500nm = 500 \times {10}^{ - 9}m \\ \beta = 0.5mm = 0.5 \times {10}^{ - 3}m \\ d = 1mm = 1 \times {10}^{ - 3}m \\ D = .........????$

λ = wavelength

β = fringe width

d = distance of two slits

D= distance between the screen and the slits.

so, we know that⬇⬇⬇⬇⬇⬇

◽the expression for the fringe width in young's double slit experiment is....

$β = \frac{Dλ}{d}$

◽here for D...⬇⬇⬇

$D = \frac{βd}{λ}$

◽we get,⬇⬇⬇

$D = \frac{0.5 \times {10}^{ - 3 \times} \times 1 \times {10}^{ - 3} }{500 \times {10}^{ - 9} } = 1 \times {10}^{ - 3}m \\ D \: = 1mm.$

so,

➡the distance between the screen and the slits is 1mm.⬅

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Answer 2

Are you preparing for competition?????????

I mean your questions tell me that either you are a medico or an engineering student