Question

diffraction pattern due to a single slit 0.1 mm wide is formed by a lens of focal length 40 cm calculate the distance between the first ARC and the next write French the wavelength of light used is 5890 A°​

Answer 1

Answer:

696hu5.mm m..m .h u have a 4756.m.m

Answer 2

Given that,

Width of slit =0.1 mm

Focal length = 40 cm

Wave length = 5890 A°

We need to calculate the distance for minima

Using formula of diffraction

$a\sin\theta=n\lambda$....(I)

We know that,

$\sin\theta=\dfrac{x}{f}$

Put the value of $\sin\theta$ in equation (I)

$a\times\dfrac{x_{1}}{f}=n\lambda$

$x_{1}=\dfrac{n\lambda f}{a}$

Put the value into the formula

$x_{1}=\dfrac{5890\times10^{-10}\times40\times10^{-2}}{0.1\times10^{-3}}$

$x_{1}= 0.002356\ m$

$x_{1}=2.36\times10^{-3}\ m$

For maxima,

$a\sin\theta=(2n+1)\times\dfrac{\lambda}{2}$

$a\times\dfrac{x_{2}}{f}=(2n+1)\times\dfrac{\lambda}{2}$

Here, n = 1

$x_{2}=\dfrac{3\lambda f}{2a}$

Put the value into the formula

$x_{2}=\dfrac{3\times5890\times10^{-10}\times40\times10^{-2}}{0.1\times10^{-3}\times2}$

$x_{2}= 0.003534\ m$

$x_{2}=3.53\times10^{-3}\ m$

We need to calculate the distance between the first dark and the next bright fringe

Using formula of distance

$x=x_{2}-x_{1}$

Put the value in to the formula

$x=3.53\times10^{-3} -2.36\times10^{-3}$

$x=0.00117\ m$

$x=1.17\times10^{-3}\ m$

Hence, The distance between the first dark and the next bright fringe is $1.17\times10^{-3}\ m$

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Topic : diffraction

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