In a Young-s double slit experiment the light source is at distance l 1- 20 -m and l 2- 40 -m from the slit-The light of wavelength -500 nm incident on slits separated at a distance d - 10 -m- A screen is placed at a distance D - 2 m away from the slits as shown in the figure- If 10k -1 maxima appears on the screen- then find the value of k- Round off your answer to the nearest integer- if required-

Path difference d sinθ=[Δ x (l_2 l_1)] sinθ =1d[nλ (l_2 l_1)] sinθ = 110× 10^ 6[n× 500× 10^ 9 20× 10^ 6] sinθ =2 [ n40 1 ] Hence,θ = sin^ 1 [ 2 ( n40 1 ) ] ...

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Standard XII

Physics

Lateral Displacement

In a Young’s ...

Question

In a Young’s double slit experiment the light source is at distance $l_{1} = 20 μ m$ and $l_{2} = 40 μ m$ from the slit.The light of wavelength $λ = 500 nm$
incident on slits separated at a distance $d = 10 μ m$ . A screen is placed at a distance $D = 2 m$ away from the slits as shown in the figure.

If $(10 k + 1)$ maxima appears on the screen, then find the value of $k$ . Round off your answer to the nearest integer, if required.

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Solution

Path difference
$d sin θ = [Δ x - (l_{2} - l_{1})]$
$sin θ = \frac{1}{d} [n λ - (l_{2} - l_{1})]$
$sin θ = \frac{1}{10 \times 10^{- 6}} [n \times 500 \times 10^{- 9} - 20 \times 10^{- 6}]$
$sin θ = 2 [\frac{n}{40} - 1]$
Hence, $θ = {sin}^{- 1} [2 (\frac{n}{40} - 1)]$
$- 1 \leq sin θ \leq 1$
$- 20 \leq [n - 40] \leq 20$
$20 \leq n \leq 60$
Hence, number of maxima $= 41$

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Path difference dsinθ=[Δx−(l2−l1)] sinθ=1d[nλ−(l2−l1)] sinθ=110×10−6[n×500×10−9−20×10−6] sinθ=2[n40−1] Hence,θ=sin−1[2(n40−1)] −1≤sinθ ≤1 −20≤[n–40]≤20 20≤n≤60 Hence, number of maxima =41