Question

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the intensity at the maximum in young's double slit experiment is Io. distance between two slits is d=5 Lambda. Lamda is is the wavelength of light used in the experiment. what will be intensity in the front of one of the slits of the screen placed at the distance D = 10 d..

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Answer 1

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Path difference

Δ=$\frac{yd}{D}$

d=5λ

D=50λ

y=$\frac{d}{2}$

=$\frac{5}{2 \gamma }$

∴Δx=$\frac{ \gamma }{4}$

Corresponding phase difference ϕ

ϕ=2πλΔx

=2πλ

=$\frac{2π}{λ}$×λ4

=$\frac{\pi}{2}$

ϕ=$\frac{ϕ2}{2}$

ϕ=$\frac{\pi}{4}$

í 1=I${cos}^{2}$$\frac{ϕ}{2}$

= í $\frac{cos}{2}$$\frac{\pi}{4}$

= í ${ \frac{1}{ \sqrt{2} } }^{2}$

=$\frac{i}{2}$

Hence a is the correct answer íѕ

$\bf\huge\boxed{\texttt{\fcolorbox{I/2}}}$.

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Answer 2

$\huge{\mathfrak{Question:-}}$The intensity at the maximum in young's double slit experiment is Io. distance between two slits is d=5 Lambda. Lamda is is the wavelength of light used in the experiment. what will be intensity in the front of one of the slits of the screen placed at the distance D = 10 d.

$\huge{\mathfrak{Answer:-}}$  

$\large{\textsf{\underline{\underline{Given:-}}}}$

$\bold{Initial\;intensity = l_{o}}$

$\bold{distance\;between\;slits = d = 5\lambda\;\;\;\; ;\lambda = wavelength\;of\;light}$

$\bold{distance\;between\;slit\;and\;screen = D = 10\times d\;\\ \\ = 10\times 5\lambda = 50\lambda}$

Here,$\bold{50\lambda}$ is the path difference, to find intensity we need phase difference which can be calculated using formula:

$\bold{Phase\;difference = \frac{2\pi }{\lambda}\times path\;difference}$

$\bold{So,\;phase\;difference = \frac{2\pi }{\lambda} \times 50 \lambda = 100\pi}$

$\bold{Now\;using\;formula\; l = \frac{l_{o}}{\frac{2cos^{2}}{2}}\;\;\;\;\; ; x = phase\;difference}$$\bold{l = \frac{l_{o}}{2cos^{2} 50\pi}}$$\bold{We\;know\;that\;cos^{2} 50\pi = 1}$

$\bold{\boxed{\boxed{{\bold {So,\; l = \frac{l_{o}}{2}}}}}}$