We have certain modification of YDSE apparatus/original YDSE apparatus. The value of intensity at point P is n I0. Find n. Given: d=1 mm , D=1 m , λ=500 nm for all cases. Intensity due to either slit on screen =I0 . P is centre of screen

Question

We have certain modification of YDSE apparatus/original YDSE apparatus. The value of intensity at point P is n I0. Find n. Given: d=1 mm , D=1 m , λ=500 nm for all cases. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/81fc2859a64cf14306557a0b2f36e0b5-.png" /> Intensity due to either slit on screen =I0 . P is centre of screen (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

I =4 I0 cos 2((π/λ) × d × 0.02) =4 I0 cos 2((π × 10-3 × 0.02/500 × 10-9)) =4 I0 cos 2(40 π)=4 I0

Q. We have certain modification of YDSE apparatus/original YDSE apparatus. The value of intensity at point $P$ is $n I_{0}$ . Find $n$ . Given: $d = 1 mm, D = 1 m, λ = 500 nm$ for all cases.

Intensity due to either slit on screen $= I_{0} . P$ is centre of screen

$I = 4 I_{0} cos^{2} (\frac{π}{λ} \times d \times 0.02)$
$= 4 I_{0} cos^{2} (\frac{π \times 1 0 ^{- 3} \times 0.02}{500 \times 1 0 ^{- 9}})$
$= 4 I_{0} cos^{2} (40 π) = 4 I_{0}$

Q. We have certain modification of YDSE apparatus/original YDSE apparatus. The value of intensity at point P is nI0​. Find n. Given: d=1mm,D=1m,λ=500nm for all cases. Intensity due to either slit on screen =I0​.P is centre of screen

I=4I0​cos2(λπ​×d×0.02) =4I0​cos2(500×10−9π×10−3×0.02​) =4I0​cos2(40π)=4I0​

Q. We have certain modification of YDSE apparatus/original YDSE apparatus. The value of intensity at point $P$ is $n I_{0}$ . Find $n$ . Given: $d = 1 mm, D = 1 m, λ = 500 nm$ for all cases.

Intensity due to either slit on screen $= I_{0} . P$ is centre of screen

$I = 4 I_{0} cos^{2} (\frac{π}{λ} \times d \times 0.02)$
$= 4 I_{0} cos^{2} (\frac{π \times 1 0 ^{- 3} \times 0.02}{500 \times 1 0 ^{- 9}})$
$= 4 I_{0} cos^{2} (40 π) = 4 I_{0}$