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Calculate the wavelength of light used in an interference experiment from the following data:
Fringe width $=0.03 \mathrm{~cm}$. Distance between the slits and eyepiece through which the interference pattern is observed is $1 \mathrm{~m}$. Distance between the images of the virtual source when a convex lens of focal length $16 \mathrm{~cm}$ is used at a distance of $80 \mathrm{~cm}$ from the eyepiece is $0.8 \mathrm{~cm}$.
Options:
Fringe width $=0.03 \mathrm{~cm}$. Distance between the slits and eyepiece through which the interference pattern is observed is $1 \mathrm{~m}$. Distance between the images of the virtual source when a convex lens of focal length $16 \mathrm{~cm}$ is used at a distance of $80 \mathrm{~cm}$ from the eyepiece is $0.8 \mathrm{~cm}$.
Solution:
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Verified Answer
The correct answer is:
$6000 Ã…$
Given: fringe with $\beta=0.03 \mathrm{~cm}, \mathrm{D}=1 \mathrm{~m}=100$
$\mathrm{cm}$
Distance between images of the source $=$ $0.8 \mathrm{~cm} .$
Image distance $v=80 \mathrm{~cm}$ Object distance $=u$ Using mirror formula,
$$
\begin{array}{l}
\qquad \begin{array}{l}
\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{60}+\frac{1}{u}=\frac{1}{16} \\
\Rightarrow \mathrm{u}=20 \mathrm{~cm}
\end{array} \\
\text { Magnification, } \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{80}{20}=4 \\
\text { Magnification } \begin{aligned}
{\text { distances between images of slits }}{\text { distance between slits }} \\
=\frac{0.8}{\mathrm{~d}}=\frac{0.8}{\mathrm{~d}}=4 \Rightarrow \mathrm{d}=0.2 \mathrm{~cm}
\end{aligned} \\
\text { Fringe width } \beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}
\end{array}
$$
or, $\beta=\frac{100 \lambda}{0.2}=0.03 \times 10^{-2}$
Therefore, wavelength of light used $\lambda=6000$ $Ã…$
$\mathrm{cm}$
Distance between images of the source $=$ $0.8 \mathrm{~cm} .$
Image distance $v=80 \mathrm{~cm}$ Object distance $=u$ Using mirror formula,
$$
\begin{array}{l}
\qquad \begin{array}{l}
\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{60}+\frac{1}{u}=\frac{1}{16} \\
\Rightarrow \mathrm{u}=20 \mathrm{~cm}
\end{array} \\
\text { Magnification, } \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{80}{20}=4 \\
\text { Magnification } \begin{aligned}
{\text { distances between images of slits }}{\text { distance between slits }} \\
=\frac{0.8}{\mathrm{~d}}=\frac{0.8}{\mathrm{~d}}=4 \Rightarrow \mathrm{d}=0.2 \mathrm{~cm}
\end{aligned} \\
\text { Fringe width } \beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}
\end{array}
$$
or, $\beta=\frac{100 \lambda}{0.2}=0.03 \times 10^{-2}$
Therefore, wavelength of light used $\lambda=6000$ $Ã…$
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